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23 tháng 4 2022

1 tháng 5 2022

ĐKXĐ: \(x\ne\pm1;x\ne0\)

a)\(\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right):\dfrac{2x}{5x-5}-\dfrac{x^2-1}{x^2+2x+1}\)

\(=\left(\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{2x}{5x-5}-\dfrac{x^2-1}{x^2+2x+1}\)

\(=\dfrac{x^2+2x+1-\left(x^2-2x+1\right)}{\left(x-1\right)\left(x+1\right)}:\dfrac{2x}{5x-5}-\dfrac{x^2-1}{x^2+2x+1}\)

\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{2x}{5x-5}-\dfrac{x^2-1}{x^2+2x+1}\)

\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}.\dfrac{5\left(x-1\right)}{2x}-\dfrac{x^2-1}{x^2+2x+1}\)

\(=\dfrac{10}{x+1}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)^2}\)

\(=\dfrac{10}{x+1}-\dfrac{x-1}{x+1}\)

\(=\dfrac{11-x}{x+1}\)

b) \(A=\dfrac{11-x}{x+1}=2\)

\(\Leftrightarrow11-x=2\left(x+1\right)\)

\(\Leftrightarrow11-x=2x+2\)

\(\Leftrightarrow-x-2x=2-11\)

\(\Leftrightarrow-3x=-9\)

\(\Leftrightarrow x=3\left(nhận\right)\)

c) -Để \(A=\dfrac{11-x}{x+1}\in Z\) thì:

\(\left(11-x\right)⋮\left(x+1\right)\)

\(\Rightarrow\left(12-x-1\right)⋮\left(x+1\right)\)

\(\Rightarrow12⋮\left(x+1\right)\)

\(\Rightarrow\left(x+1\right)\inƯ\left(12\right)\)

\(\Rightarrow\left(x+1\right)\in\left\{1;2;3;4;6;12;-1;-2;-3;-4;-6;-12\right\}\)

\(\Rightarrow x\in\left\{2;3;5;11;-2;-3;-4;-5;-7;-13\right\}\)

 

 

 

1 tháng 5 2022

em cảm ưn gất nhìuuuuu:33

25 tháng 2 2022

\(A=\left(\dfrac{1}{x^2-1}+\dfrac{1}{x+1}\right):\left(\dfrac{1}{x-1}-\dfrac{1}{x}\right)\)

\(\Rightarrow A=\left(\dfrac{1}{\left(x-1\right)\left(x+1\right)}+\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\right):\left(\dfrac{x}{x\left(x-1\right)}-\dfrac{x-1}{x\left(x-1\right)}\right)\)

\(\Rightarrow A=\dfrac{1+x-1}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-x+1}{x\left(x-1\right)}\)

\(\Rightarrow A=\dfrac{x}{\left(x-1\right)\left(x+1\right)}:\dfrac{1}{x\left(x-1\right)}\)

\(\Rightarrow A=\dfrac{x}{\left(x-1\right)\left(x+1\right)}.x\left(x-1\right)\)

\(\Rightarrow A=\dfrac{x^2}{x+1}\)

25 tháng 2 2022

đk : xkhác -1 ; 1 

\(A=\left(\dfrac{1+x-1}{\left(x+1\right)\left(x-1\right)}\right):\left(\dfrac{x-x+1}{x\left(x-1\right)}\right)=\dfrac{x}{\left(x+1\right)\left(x-1\right)}:\dfrac{1}{x\left(x-1\right)}=\dfrac{x^2}{x+1}\)

Câu 1: 

1: Ta có: \(P=\left(\dfrac{x^2}{x^2-3}+\dfrac{2x^2-24}{x^4-9}\right)\cdot\dfrac{7}{x^2+8}\)

\(=\left(\dfrac{x^2\left(x^2+3\right)}{\left(x^2-3\right)\left(x^2+3\right)}+\dfrac{2x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\right)\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{x^4+3x^2+2x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{x^4+5x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{x^4+8x^2-3x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{x^2\left(x^2+8\right)-3\left(x^2+8\right)}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{\left(x^2+8\right)\left(x^2-3\right)}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{7}{x^2+3}\)

NV
3 tháng 4 2021

Câu 2a đề sai, pt này ko giải được

2b.

\(P\left(x\right)=\left(2x+7\right)\left(x^2-4x+4\right)+\left(a+20\right)x+\left(b-28\right)\)

Do \(\left(2x+7\right)\left(x^2-4x+4\right)⋮\left(x^2-4x+4\right)\)

\(\Rightarrow P\left(x\right)\) chia hết \(Q\left(x\right)\) khi \(\left(a+20\right)x+\left(b-28\right)\) chia hết \(x^2-4x+4\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+20=0\\b-28=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-20\\b=28\end{matrix}\right.\)

3a.

\(VT=\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}=\dfrac{2+x^2+y^2}{1+x^2+y^2+x^2y^2}=1+\dfrac{1-x^2y^2}{1+x^2+y^2+x^2y^2}\le1+\dfrac{1-x^2y^2}{1+2xy+x^2y^2}\)

\(VT\le1+\dfrac{\left(1-xy\right)\left(1+xy\right)}{\left(xy+1\right)^2}=1+\dfrac{1-xy}{1+xy}=\dfrac{2}{1+xy}\) (đpcm)

3b

Ta có: \(n^3-n=n\left(n-1\right)\left(n+1\right)\) là tích 3 số nguyên liên tiếp nên luôn chia hết cho 6

\(\Rightarrow n^3\) luôn đồng dư với n khi chia 6

\(\Rightarrow S\equiv2021^{2022}\left(mod6\right)\)

Mà \(2021\equiv1\left(mod6\right)\Rightarrow2021^{2020}\equiv1\left(mod6\right)\)

\(\Rightarrow2021^{2022}-1⋮6\)

\(\Rightarrow S-1⋮6\)

a) Ta có: \(P=\left(\dfrac{3}{x+1}+\dfrac{x-9}{x^2-1}+\dfrac{2}{1-x}\right):\dfrac{x-3}{x^2-1}\)

\(=\left(\dfrac{3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{x-9}{\left(x+1\right)\left(x-1\right)}-\dfrac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{x-3}{x^2-1}\)

\(=\dfrac{3x-3+x-9-2x-2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{x-3}\)

\(=\dfrac{2x-14}{x-3}\)

b) Ta có: \(x^2-9=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(loại\right)\\x=-3\left(nhận\right)\end{matrix}\right.\)

Thay x=-3 vào biểu thức \(P=\dfrac{2x-14}{x-3}\), ta được:

\(P=\dfrac{2\cdot\left(-3\right)-14}{-3-3}=\dfrac{-20}{-6}=\dfrac{10}{3}\)

Vậy: Khi \(x^2-9=0\) thì \(P=\dfrac{10}{3}\)

c) Để P nguyên thì \(2x-14⋮x-3\)

\(\Leftrightarrow2x-6-8⋮x-3\)

mà \(2x-6⋮x-3\)

nên \(-8⋮x-3\)

\(\Leftrightarrow x-3\inƯ\left(-8\right)\)

\(\Leftrightarrow x-3\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)

\(\Leftrightarrow x\in\left\{4;2;5;1;7;-1;11;-5\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{4;2;5;7;11;-5\right\}\)

Vậy: Để P nguyên thì \(x\in\left\{4;2;5;7;11;-5\right\}\)

22 tháng 10 2023

a) \(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\left(x>0;x\ne1\right)\)

\(=\dfrac{x\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x\sqrt{x}-2x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}\left(x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\)

\(=\sqrt{x}-1\)

\(---\)

b) Thay \(x=3+2\sqrt{2}\) vào \(A\), ta được:

\(A=\sqrt{3+2\sqrt{2}}-1\)

\(=\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot1+1^2}-1\)

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

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

\(=\sqrt{2}+1-1\)

\(=\sqrt{2}\)

\(Toru\)

22 tháng 10 2023

\(a,A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\left(dk:x>0,x\ne1\right)\\ =\dfrac{x}{\sqrt{x}-1}-\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{x}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\\ =\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\\ =\sqrt{x}-1\)

\(b,x=3+2\sqrt{2}=\sqrt{2}^2+2\sqrt{2}.1+1=\left(\sqrt{2}+1\right)^2\)

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

a) ĐKXĐ: \(x\notin\left\{0;-\dfrac{1}{2};\dfrac{1}{2}\right\}\)

Ta có: \(A=\left(\dfrac{1}{2x-1}+\dfrac{3}{1-4x^2}-\dfrac{2}{2x+1}\right):\left(\dfrac{x^2}{2x^2+x}\right)\)

\(=\left(\dfrac{2x+1}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{3}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{2\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}\right):\left(\dfrac{x^2}{x\left(2x+1\right)}\right)\)

\(=\dfrac{2x+1-3-4x+2}{\left(2x-1\right)\left(2x+1\right)}:\dfrac{x}{2x+1}\)

\(=\dfrac{-2x}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{2x+1}{x}\)

\(=\dfrac{-2}{2x-1}\)

b) Ta có: \(\left|2x-1\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\left(nhận\right)\\x=-\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)

Thay \(x=\dfrac{3}{2}\) vào biểu thức \(A=\dfrac{-2}{2x-1}\), ta được:

\(A=-2:\left(2\cdot\dfrac{3}{2}-1\right)=-2:\left(3-1\right)=-2:2=-1\)

Vậy: Khi \(\left|2x-1\right|=2\) thì A=-1

c) Để \(A=\dfrac{1}{3}\) thì \(\dfrac{-2}{2x-1}=\dfrac{1}{3}\)

\(\Leftrightarrow2x-1=-6\)

\(\Leftrightarrow2x=-5\)

hay \(x=-\dfrac{5}{2}\)(thỏa ĐK)

Vậy: Để \(A=\dfrac{1}{3}\) thì \(x=-\dfrac{5}{2}\)

4 tháng 2 2021

Cảm ơn bạn nhiều ạ!

a: Thay x=5 vào B, ta được:

\(B=\dfrac{5-1}{5-3}=\dfrac{4}{2}=2\)

b:  \(A=\dfrac{2x^2+6x-2x^2-3x-1}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-1}{\left(x+3\right)\left(x-3\right)}\)

Ta có: \(A=\dfrac{2x}{1-x^3}+\dfrac{1}{x^2-x}+\dfrac{1}{x^2+x+1}\)

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

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

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

Thay x=10 vào A, ta được:

\(A=\dfrac{1}{10\cdot\left(10^3-1\right)}=\dfrac{1}{10\cdot999}=\dfrac{1}{9990}\)