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Những câu trả lời của 💋ɯɐN ɥu∀ ƃuàoH💋:

Vào lúc: 2019-09-17 20:18:42 Xem câu hỏi

put the verbs in brackets in the correct tense form

1. School (finish) ......finishes..........at 4.30 p.m every day.

2. .....Do.........you often (eat) ......eat.............lunch in the school canteen?

3.-Where (be) .....are............you ,Phong? - I'm upstais .I (do) ......am doing..............my homework.

4.it's warm today . I (not want ) ...........don't want................(stay) .......to stay..................at home now . What about (go)............going...........swimming in the river?

5. Listen! ..........Are...........they (sing ) ..........singing...........in the classroom?

6.My family s (spend).........spending...............out summer holidays at the seaside.

Vào lúc: 2019-09-17 20:12:24 Xem câu hỏi

\(\left(\sqrt{10}+\sqrt{2}\right)\sqrt{3-\sqrt{5}}=\sqrt{2}\left(\sqrt{5}+1\right)\sqrt{3-\sqrt{5}}=\left(\sqrt{5}+1\right)\sqrt{6-2\sqrt{5}}=\left(\sqrt{5}+1\right)\sqrt{\left(\sqrt{5}-1\right)^2}\)\(=\left(\sqrt{5}+1\right)|\sqrt{5}-1|=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=\sqrt{5}^2-1^2=4\)

Vào lúc: 2019-09-15 22:47:36 Xem câu hỏi

\(a,25-x^4=5^2-\left(x^2\right)^2=\left(5-x^2\right)\left(5+x^2\right)\)

\(b,\left(3x+y\right)^2=\left(3x\right)^2+2.3x.y+y^2=9x^2+6xy+y^2\)

\(c,\left(x+1\right)^2=x^2+2x+1\)

\(d,\left(x+3\right)\left(x^2-3x+9\right)=x^3+27\)

\(e,\left(2+3x\right)^2=2^2+2.2.3x+\left(3x\right)^2=4+12x+9x^2\)

\(f,x^2-9=\left(x-3\right)\left(x+3\right)\)

\(g,\left(x-1\right)^3=x^3-3.x^2.1+3.x.1^2-1^3=x^3-3x^2+3x-1\)

\(h,x^3-8=\left(x-2\right)\left(x^2-2x+4\right)\)

Vào lúc: 2019-09-13 14:44:48 Xem câu hỏi

\(x^2-4x+5\)

\(\Delta'=\left(-2\right)^2-1.5=4-5=-1< 0\)

=> PT vô nghiệm

\(x^2+x+\frac{1}{2}\)

\(\Delta=1^2-4.1.\frac{1}{2}=1-2=-1< 0\)

=>PT vô nghiệm

Vào lúc: 2019-08-05 13:17:41 Xem câu hỏi

\(x-\sqrt{2x-9}=6\left(Đk:x\ge\frac{9}{2}\right)\)

\(\Rightarrow\sqrt{2x-9}=x-6\left(đk:x\ge6\right)\)

\(\Rightarrow\left(\sqrt{2x-9}\right)^2=\left(x-6\right)^2\)

\(\Rightarrow2x-9=x^2-12x+36\)

\(\Rightarrow-x^2+14x-45=0\)

\(\Rightarrow-x^2+9x+5x-45=0\)

\(\Rightarrow-x\left(x-9\right)+5\left(x-9\right)=0\)

\(\Rightarrow\left(x-9\right)\left(5-x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=9\\x=5\end{cases}\left(TM\right)}\)

Vào lúc: 2019-08-04 10:07:17 Xem câu hỏi

\(A=\frac{3}{x^4-x^3+x-1}-\frac{1}{x^4+x^3-x-1}-\frac{4}{x^5-x^4+x^3-x^2+x-1}\)

\(=\frac{3}{\left(x-1\right)\left(x^3+1\right)}-\frac{1}{\left(x+1\right)\left(x^3-1\right)}-\frac{4}{\left(x-1\right)\left(x^4+x^2+1\right)}\)

\(=\frac{3}{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}-\frac{1}{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}-\frac{4}{\left(x-1\right)\left(x^4+x^2+1\right)}\)

\(=\left[\frac{3}{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}-\frac{1}{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}\right]-\frac{4}{\left(x-1\right)\left(x^4+x^2+1\right)}\)

\(=\left[\frac{3\left(x^2+x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}-\frac{x^2-x+1}{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\right]-\frac{4}{\left(x-1\right)\left(x^4+x^2+1\right)}\)\(=\frac{3x^2+3x+3-x^2+x-1}{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}-\frac{4}{\left(x-1\right)\left(x^4+x^2+1\right)}\)

\(=\frac{2x^2+4x+2}{\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)}-\frac{4}{\left(x-1\right)\left(x^4+x^2+1\right)}\)

\(=\frac{2x^2+4x+2}{\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)}-\frac{4\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)}\)

\(=\frac{2x^2+4x+2-4x-4}{\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)}=\frac{2x^2-2}{\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)}=\frac{2\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)}\)

\(=\frac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)}=\frac{2}{x^4+x^2+1}\)

\(\Rightarrow A=\frac{2}{x^4+x^2+1}\left(x\ne\pm1\right)\)

Ta có: \(x^4+x^2+1=\left(x^2\right)^2+2.x^2.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x^2+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)

Vậy A > 0 với mọi \(x\ne\pm1\)

Vào lúc: 2019-08-04 09:02:09 Xem câu hỏi

\(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x+\sqrt{x^2+1}}=y+\sqrt{y^2+1}\\\frac{1}{y+\sqrt{y^2+1}}=x+\sqrt{x^2+1}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-x+\sqrt{x^2+1}=y+\sqrt{y^2+1}\left(1\right)\\-y+\sqrt{y^2+1}=x+\sqrt{x^2+1}\left(2\right)\end{cases}}\)

Cộng vế với vế của (1) và (2) ta có:

\(-2x-2y=0\Leftrightarrow-2\left(x+y\right)=0\Leftrightarrow x+y=0\)

\(\Rightarrow P=x^{2019}+y^{2019}=0\)

Vào lúc: 2019-08-02 22:03:09 Xem câu hỏi

Ta có:\(a+b\ge2\sqrt{ab}\Rightarrow a-2\sqrt{ab}+b\ge0\Rightarrow\left(\sqrt{a}\right)^2-2\sqrt{ab}+\left(\sqrt{b}\right)^2\ge0\Rightarrow\left(\sqrt{a}+\sqrt{b}\right)^2\ge0\left(LĐ\right)\)Dấu "=" xảy ra <=> a = b

Vào lúc: 2019-08-02 08:34:13 Xem câu hỏi

\(a,|x+3|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}x+3=\frac{1}{2}\\x+3=\frac{-1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-5}{2}\\x=\frac{-7}{2}\end{cases}}\)

\(b,|2x+3|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}2x+3=\frac{1}{2}\\2x+3=\frac{-1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-4}{3}\\x=\frac{-7}{4}\end{cases}}\)

\(c,2x+3=0\Leftrightarrow2x=-3\Leftrightarrow x=\frac{-3}{2}\)

\(d,|2x+3|-1=\frac{1}{2}\Leftrightarrow|2x+3|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}2x+3=\frac{3}{2}\\2x+3=\frac{-3}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-3}{4}\\x=\frac{-9}{4}\end{cases}}\)

\(e,|2x+3|+5=\frac{1}{2}\Leftrightarrow|2x+3|=\frac{-9}{2}\)(vô lí)

\(f,4-|2x+3|=1\Leftrightarrow|2x+3|=3\Leftrightarrow\orbr{\begin{cases}2x+3=3\\2x+3=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

Vào lúc: 2019-08-01 22:55:01 Xem câu hỏi

\(x^2-2\left(m-1\right)x+m-3=0\)

\(\Delta'=b'^2-ac=\left[-\left(m-1\right)^2\right]-1.\left(m-3\right)=m^2-2m+1-m+3=m^2-3m+4\)

\(=m^2-2.m.\frac{3}{2}+\frac{9}{4}+\frac{7}{4}=\left(m-\frac{3}{2}\right)^2+\frac{7}{4}\)

\(\left(m-\frac{3}{2}\right)^2\ge0\forall m\)

\(\Rightarrow\left(m-\frac{3}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}>0\forall m\)

\(\Rightarrow\Delta'>0\forall m\)

Vậy...

Vào lúc: 2019-07-31 21:36:05 Xem câu hỏi

\(a,\)Vì \(a< b\Rightarrow a-b< 0\)

\(\Leftrightarrow\sqrt{a}^2-\sqrt{b}^2< 0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)< 0\)

Mà \(a,b>0\Rightarrow\sqrt{a}+\sqrt{b}>0\)

\(\Rightarrow\sqrt{a}-\sqrt{b}< 0\)

\(\Rightarrow\sqrt{a}< \sqrt{b}\left(đpcm\right)\)

\(b,\)Ta có:\(a\ge0;b>0\Rightarrow\sqrt{a}+\sqrt{b}>0\)

\(\sqrt{a}< \sqrt{b}\Rightarrow\sqrt{a}-\sqrt{b}< 0\)(1)

Nhân hai vế của (1) với \(\sqrt{a}+\sqrt{b}\).Mà theo cmt thì \(\sqrt{a}+\sqrt{b}>0\)nên khi nhân vào thì dấu của BPT (1) không đổi chiều

\(\Rightarrow\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)< 0\left(\sqrt{a}+\sqrt{b}\right)\)

\(\Leftrightarrow\sqrt{a}^2-\sqrt{b}^2< 0\)

\(\Leftrightarrow a-b< 0\)

\(\Rightarrow a< 0\left(đpcm\right)\)

Vào lúc: 2019-07-30 17:46:23 Xem câu hỏi

Ta có:\(B=\frac{\sqrt{x}+1}{\sqrt{x}-3}\left(Đk:x\ne9\right)=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)

Để B nhận giá trị nguyên \(\Leftrightarrow4⋮\sqrt{x}-3\)

Vì \(x\in Z\Rightarrow\sqrt{x}-3\inƯ_{\left(4\right)}=\left\{\pm1;\pm2\pm4\right\}\)

Ta có bảng sau:

\(\sqrt{x}-3\)\(1\)\(-1\)\(2\)\(-2\)\(4\)\(-4\)
\(x\)\(2\)(TM)\(4\)(TM)\(25\)(TM)\(1\)(TM)\(49\)(TM)KTM

Vào lúc: 2019-07-30 16:38:12 Xem câu hỏi

\(a,\left(1+\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\left(1-\frac{a+\sqrt{a}}{1+\sqrt{a}}\right)=\left(1+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)

\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1^2-\sqrt{a}^2=1-a\)

\(b,\left(2-\frac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(2-\frac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)=\left(2-\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right)\left(2-\frac{-\sqrt{a}\left(\sqrt{b}-5\right)}{\sqrt{b}-5}\right)\)

\(=\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)=2^2-\sqrt{a}^2=2-a\)

\(c,\left(3+\frac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(3-\frac{3a+\sqrt{a}}{3\sqrt{a}+1}\right)=\left(3+\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{\sqrt{a}-2}\right)\left(3-\frac{\sqrt{a}\left(3\sqrt{a}+1\right)}{3\sqrt{a}+1}\right)\)

\(=\left(3+\sqrt{a}\right)\left(3-\sqrt{a}\right)=3^2-\sqrt{a}^2=3-a\)

\(d,\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}+2\right)\left(2-\frac{\sqrt{a}+a}{1+\sqrt{a}}\right)=\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}+2\right)\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)

\(=\left(\sqrt{a}+2\right)\left(2-\sqrt{a}\right)=2^2-\sqrt{a}^2=2-a\)

Vào lúc: 2019-07-30 10:32:33 Xem câu hỏi

\(3\left(x+2\right)+2x+4=3\left(x+2\right)+2\left(x+2\right)=5\left(x+2\right)\)

\(4\left(x-2\right)-x+2=4\left(x-2\right)-\left(x-2\right)=3\left(x-2\right)\)

Vào lúc: 2019-07-29 12:15:01 Xem câu hỏi

\(2x^2-5x+2=2x^2-4x-x+2=2x\left(x-2\right)-\left(x-2\right)=\left(x-2\right)\left(2x-1\right)\)

Vào lúc: 2019-07-29 08:13:01 Xem câu hỏi

\(a,A=\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{3}{\sqrt{x}+2}-\frac{9\sqrt{x}-10}{x-4}\left(x\ge0;x\ne16\right)\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{9\sqrt{x}-10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{3\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{9\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{x+2\sqrt{x}+3\sqrt{x}-6-9\sqrt{x}+10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{x-4\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\left(\sqrt{x}\right)^2-2.\sqrt{x}.2+2^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-2}{\sqrt{x}+2}\)

Vây...

\(b,\)Ta có:\(x=4-2\sqrt{3}=\left(1-\sqrt{3}\right)^2\)

Thay \(x=\left(1-\sqrt{3}\right)^2\)vào A ta được:

\(A=\frac{\sqrt{\left(1-\sqrt{3}\right)^2}-2}{\sqrt{\left(1-\sqrt{3}\right)^2}+2}=\frac{\sqrt{3}-1-2}{\sqrt{3}-1+2}=\frac{\sqrt{3}-3}{\sqrt{3}-1}=\frac{-\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}=-\sqrt{3}\)

Vào lúc: 2019-07-28 22:51:35 Xem câu hỏi

\(x^4-8x^3+11x^2+8x-12=0\)

\(\Leftrightarrow x^4-x^3-7x^3+7x^2+4x^2-4x+12x-12=0\)

\(\Leftrightarrow\left(x^4-x^3\right)-\left(7x^3-7x^2\right)+\left(4x^2-4x\right)+\left(12x-12\right)=0\)

\(\Leftrightarrow x^3\left(x-1\right)-7x^2\left(x-1\right)+4x\left(x-1\right)+12\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-7x^2+4x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2-8x^2-8x+12x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)[x^2\left(x+1\right)-8x\left(x+1\right)+12\left(x+1\right)]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^2-8x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x-6\right)=0\)

\(\Leftrightarrow\)x - 1 =0 ; x + 1 = 0 ; x - 2 =0 hoặc x - 6 = 0

\(\Leftrightarrow\)x = 1 ; x = -1 ; x = 2 ; x=6

Vào lúc: 2019-07-28 10:11:38 Xem câu hỏi

\(M=\frac{x+\sqrt{x^2-2x}}{x-\sqrt{x^2-2x}}-\frac{x-\sqrt{x^2-2x}}{x+\sqrt{x^2-2x}}\left(x< 0;x\ge2\right)\)

\(=\frac{\left(x+\sqrt{x^2-2x}\right)\left(x+\sqrt{x^2-2x}\right)}{x^2-\sqrt{x^2-2x}^2}-\frac{\left(x-\sqrt{x^2-2x}\right)\left(x-\sqrt{x^2-2x}\right)}{x^2-\sqrt{x^2-2x}^2}\)

\(=\frac{x^2+x\sqrt{x^2-2x}+x\sqrt{x^2-2x}+x^2-2x}{x^2-x^2-2x}-\frac{x^2-x\sqrt{x^2-2x}-x\sqrt{x^2-2x}+x^2-2x}{x^2-x^2-2x}\)

\(=\frac{2x^2+2x\sqrt{x^2-2x}-2x}{-2x}-\frac{2x^2-2\sqrt{x^2-2x}-2x}{-2x}\)

\(=\frac{2x^2+2x\sqrt{x^2-2x}-2x-2x^2+2x\sqrt{x^2-2x}+2x}{-2x}\)

\(=\frac{4x\sqrt{x^2-2x}}{-2x}=-2x\sqrt{x^2-2x}\)

Vào lúc: 2019-07-28 09:52:42 Xem câu hỏi

Chỉ làm thử thôi nhé-.-

\(B=\left(\sqrt{x+2-4\sqrt{x-2}}+\sqrt{x+2+4\sqrt{x-2}}\right):\sqrt{\frac{4}{x^2}-\frac{4}{x}+1}\left(đk:x\ge2\right)\)

\(=\left(\sqrt{x-2-2\sqrt{x-2}.2+2^2}+\sqrt{x-2+2\sqrt{x-2}.2+2^2}\right):\sqrt{\frac{4}{x^2}-\frac{4x}{x^2}+\frac{x^2}{x^2}}\)

\(=[\left(\sqrt{\left(\sqrt{x-2}-2\right)^2}+\sqrt{\left(\sqrt{x-2}+2\right)^2}\right):\sqrt{\frac{4-4x+x^2}{x^2}}\)

\(=\left(|\sqrt{x-2}-2|+|\sqrt{x-2}+2|\right):\sqrt{\frac{\left(2-x\right)^2}{x^2}}\)

\(=\left(\sqrt{x-2}-2+\sqrt{x-2}+2\right).\frac{x}{2-x}\)

\(=2\sqrt{x-2}.\frac{x}{2-x}=\frac{2x\sqrt{x-2}}{2-x}\)

Vào lúc: 2019-07-28 09:20:01 Xem câu hỏi

\(a,E=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\left(Đk:x\ge0;x\ne\pm1\right)\)(Đề như này mới đúng!)

\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{-\left(3x+9\sqrt{x}-2\sqrt{x}-6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{2x-2\sqrt{x}+3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{7\sqrt{x}-2-5x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{5\sqrt{x}+2\sqrt{x}-2-5x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\left(5\sqrt{x}-5x\right)+\left(2\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)

Vậy...

\(b,\)Ta có:\(\frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{-15+17-5\sqrt{x}}{\sqrt{x}+3}=\frac{\left(-15-5\sqrt{x}\right)+17}{\sqrt{x}+3}=\frac{-5\left(\sqrt{x}+3\right)+17}{\sqrt{x}+3}=-5+\frac{17}{\sqrt{x}+3}\)

Vì \(\sqrt{x}\ge0\forall x\Rightarrow\sqrt{x}+3\ge3\forall x\Rightarrow\frac{17}{\sqrt{x}+3}\le\frac{17}{3}\Rightarrow-5+\frac{17}{\sqrt{x}+3}\le\frac{2}{3}\Rightarrow E\le\frac{2}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

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