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a: \(x^2\cdot2\sqrt{3}+x+1=\sqrt{3}\cdot\left(x+1\right)\)
=>\(x^2\cdot2\sqrt{3}+x\left(1-\sqrt{3}\right)+1-\sqrt{3}=0\)
\(\text{Δ}=\left(1-\sqrt{3}\right)^2-4\cdot2\sqrt{3}\left(1-\sqrt{3}\right)\)
\(=4-2\sqrt{3}-8\sqrt{3}\left(1-\sqrt{3}\right)\)
\(=4-2\sqrt{3}-8\sqrt{3}+24=28-10\sqrt{3}=\left(5-\sqrt{3}\right)^2>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}x_1=\dfrac{-\left(1-\sqrt{3}\right)-\left(5-\sqrt{3}\right)}{2\cdot2\sqrt{3}}=\dfrac{-1+\sqrt{3}-5+\sqrt{3}}{4\sqrt{3}}=\dfrac{1-\sqrt{3}}{2}\\x_2=\dfrac{-\left(1-\sqrt{3}\right)+5-\sqrt{3}}{2\cdot2\sqrt{3}}=\dfrac{4}{4\sqrt{3}}=\dfrac{1}{\sqrt{3}}\end{matrix}\right.\)
b: \(5x^2-3x+1=2x+31\)
=>\(5x^2-3x+1-2x-31=0\)
=>\(5x^2-5x-30=0\)
=>\(x^2-x-6=0\)
=>(x-3)(x+2)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
c: \(x^2+2\sqrt{2}x+4=3\left(x+\sqrt{2}\right)\)
=>\(x^2+2\sqrt{2}x+4-3x-3\sqrt{2}=0\)
=>\(x^2+x\left(2\sqrt{2}-3\right)+4-3\sqrt{2}=0\)
\(\text{Δ}=\left(2\sqrt{2}-3\right)^2-4\left(4-3\sqrt{2}\right)\)
\(=17-12\sqrt{2}-16+12\sqrt{2}=1\)>0
Do đó, phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}x_1=\dfrac{-\left(2\sqrt{2}-3\right)-1}{2}=\dfrac{-2\sqrt{2}+3-1}{2}=-\sqrt{2}+1\\x_2=\dfrac{-\left(2\sqrt{2}-3\right)+1}{2}=\dfrac{-2\sqrt{2}+4}{2}=-\sqrt{2}+2\end{matrix}\right.\)
a. ĐKXĐ: \(-1\le x\le1\)
Đặt \(\sqrt{1+x}+\sqrt{1-x}=t>0\)
\(\Rightarrow t^2=2+2\sqrt{1-t^2}\)
Pt trở thành:
\(t.t^2=8\Leftrightarrow t^3=8\Leftrightarrow t=2\)
\(\Rightarrow\sqrt{1+x}+\sqrt{1-x}=2\)
\(\Leftrightarrow2+2\sqrt{1-x^2}=2\)
\(\Leftrightarrow1-x^2=0\Rightarrow x=\pm1\)
b.
ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)
\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\)
Pt trở thành:
\(t=t^2-4-16\Leftrightarrow...\)
Câu 1:
\(A=21\left(a+\frac{1}{b}\right)+3\left(b+\frac{1}{a}\right)=21a+\frac{21}{b}+3b+\frac{3}{a}\)
\(=(\frac{a}{3}+\frac{3}{a})+(\frac{7b}{3}+\frac{21}{b})+\frac{62}{3}a+\frac{2b}{3}\)
Áp dụng BĐT Cô-si:
\(\frac{a}{3}+\frac{3}{a}\geq 2\sqrt{\frac{a}{3}.\frac{3}{a}}=2\)
\(\frac{7b}{3}+\frac{21}{b}\geq 2\sqrt{\frac{7b}{3}.\frac{21}{b}}=14\)
Và do $a,b\geq 3$ nên:
\(\frac{62}{3}a\geq \frac{62}{3}.3=62\)
\(\frac{2b}{3}\geq \frac{2.3}{3}=2\)
Cộng tất cả những BĐT trên ta có:
\(A\geq 2+14+62+2=80\) (đpcm)
Dấu "=" xảy ra khi $a=b=3$
Câu 2:
Bình phương 2 vế ta thu được:
\((x^2+6x-1)^2=4(5x^3-3x^2+3x-2)\)
\(\Leftrightarrow x^4+12x^3+34x^2-12x+1=20x^3-12x^2+12x-8\)
\(\Leftrightarrow x^4-8x^3+46x^2-24x+9=0\)
\(\Leftrightarrow (x^2-4x)^2+6x^2+24(x-\frac{1}{2})^2+3=0\) (vô lý)
Do đó pt đã cho vô nghiệm.
a)Pt \(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\dfrac{1}{3}+\dfrac{1}{2}\)
\(\Leftrightarrow\left|2x-1\right|=\dfrac{5}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\dfrac{5}{6}\\2x-1=-\dfrac{5}{6}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{12}\\x=\dfrac{1}{12}\end{matrix}\right.\)
Vậy...
b)Đk:\(x\ge3\)
Pt \(\Leftrightarrow\sqrt{x-3}\left(x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\x-4=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=4\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)
Vậy...
c)Đk:\(x\ge1\)
\(x+\sqrt{x-1}=13\)
\(\Leftrightarrow\sqrt{x-1}=13-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}13-x\ge0\\x-1=x^2-26x+169\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\x^2-27x+170=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\x^2-17x-10x+170=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\\left(x-17\right)\left(x-10\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\\left[{}\begin{matrix}x=17\\x=10\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow x=10\) (tm)
Vậy...
a) \(\sqrt{7+\sqrt{2x}=3+\sqrt{5}}\) (x≥0) Đặt \(\sqrt{2x}\) = a ( a>0 )
Khi đó pt :
<=> 7+a =3 + \(\sqrt{5}\)
<=> 4+a = \(\sqrt{5}\)
<=> (4+a)\(^2\) = 5
<=> 16 + 8a + a\(^2\) = 5
<=>a\(^2\) + 8a+ 11 = 0
<=> a = -4 + \(\sqrt{5}\) (Loại) và a = -4-\(\sqrt{5}\)(Loại)
Vậy Pt vô nghiệm.
b) \(\sqrt{3x^2-4x}\) = 2x-3
<=> 3x\(^2\)- 4x = 4x\(^2\)-12x + 9
<=> x\(^2\)-8x+9 = 0
<=> x=1 , x=9
Vậy S={1;9}
c\(\dfrac{\left(7-x\right)\sqrt{7-x}+\left(x-5\right)\sqrt{x-5}}{\sqrt{7-x}+\sqrt{x-5}}\) = 2
<=> \(\dfrac{\left(\sqrt{7-x}\right)^3+\left(\sqrt{x-5}\right)^3}{\sqrt{7-x}+\sqrt{x-5}}=2\)
<=> \(\dfrac{\left(\sqrt{7-x}+\sqrt{x-5}\right)\left(7-x-\sqrt{\left(7-x\right)\left(x-5\right)}+x-5\right)}{\sqrt{7-x}+\sqrt{x-5}}=2\)
<=> \(\sqrt{\left(7-x\right)\left(x-5\right)}=0\)
<=> x=7,x=5
Vậy x=5 hoặc x=7
E hổng biết cách này có đúng ko nữa:((
5
Ta có:\(S=\frac{2010}{x}+\frac{1}{2010y}+\frac{1010}{1005}\ge2\sqrt{\frac{2010}{x}\cdot\frac{1}{2010y}}+\frac{1010}{1005}\left(AM-GM\right)\)
\(=\frac{2}{\sqrt{xy}}+\frac{2010}{1005}\ge\frac{2}{\frac{x+y}{2}}+2=4\)( AM-GM ngược dấu )
Dấu "=" xảy ra khi \(x=y=\frac{2010}{4024}\)
a) Ta có: \(\sqrt{\left(x+1\right)^2}=3\)
\(\Leftrightarrow\left|x+1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
b) Ta có: \(3\sqrt{4x+4}-\sqrt{9x-9}-8\sqrt{\dfrac{x+1}{16}}=5\)
\(\Leftrightarrow6\sqrt{x+1}-3\sqrt{x-3}-2\sqrt{x+1}=5\)
\(\Leftrightarrow4\sqrt{x+1}=5+3\sqrt{x-3}\)
\(\Leftrightarrow16\left(x+1\right)=25+30\sqrt{x-3}+9\left(x-3\right)\)
\(\Leftrightarrow16x+16=25+9x-27+30\sqrt{x-3}\)
\(\Leftrightarrow30\sqrt{x-3}=16x+16+2-9x\)
\(\Leftrightarrow30\sqrt{x-3}=7x+18\)
\(\Leftrightarrow x-3=\left(\dfrac{7x+18}{30}\right)^2\)
\(\Leftrightarrow x-3=\dfrac{49x^2}{900}+\dfrac{7}{25}x+\dfrac{9}{25}\)
\(\Leftrightarrow\dfrac{49}{900}x^2-\dfrac{18}{25}x+\dfrac{84}{25}=0\)
\(\Delta=\left(-\dfrac{18}{25}\right)^2-4\cdot\dfrac{49}{900}\cdot\dfrac{84}{25}=-\dfrac{16}{75}< 0\)
Vậy: Phương trình vô nghiệm
a)Pt\(\Leftrightarrow\left|x+1\right|=3\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
b)Đk:\(x\ge-1\)
Sửa đề: \(3\sqrt{4x+4}-\sqrt{9x+9}-8\sqrt{\dfrac{x+1}{16}}=5\)
Pt \(\Leftrightarrow6\sqrt{x+1}-3\sqrt{x+1}-2\sqrt{x+1}=5\)
\(\Leftrightarrow\sqrt{x+1}=5\)
\(\Leftrightarrow x=24\left(tm\right)\)
a) \(\sqrt{\left(x-3\right)^2}=2\Rightarrow\left|x-3\right|=2\Rightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b) \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Rightarrow\sqrt{9\left(x+2\right)}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25\left(x+2\right)}=6\)
\(\Rightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Rightarrow2\sqrt{x+2}=6\Rightarrow\sqrt{x+2}=3\Rightarrow x+2=9\Rightarrow x=7\)
\(Q=\dfrac{1}{x-2\sqrt{x}+3}\)
Ta có: \(x-2\sqrt{x}+3=x-2\sqrt{x}+1+2=\left(\sqrt{x}-1\right)^2+2\ge2\)
\(\Rightarrow\dfrac{1}{x-2\sqrt{x}+3}\le2\Rightarrow Q_{max}=2\) khi \(x=1\)
a/ Tớ làm sai rồi . giờ làm lại
\(x^2-2010+\sqrt{x+2010}=0\Rightarrow-\left(x+2010\right)+\sqrt{x+2010}+x^2+x=0\)
Đặt \(a=\sqrt{x+2010}\left(a\ge0\right)\) ta được pt: -a2 + a + x2 + x = 0
Có: \(\Delta=1+4x^2+4x=\left(2x+1\right)^2\Rightarrow\sqrt{\Delta}=2x+1\)
\(\Rightarrow\orbr{\begin{cases}a=-x\left(l\right)\\a=x+1\left(n\right)\end{cases}}\)
Với a = x + 1 \(\Rightarrow\sqrt{x+2010}=x+1\Rightarrow x+2010=\left(x+1\right)^2\). Tới đây bạn tự làm
Sory em mới học lớp 7