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a,ĐKXĐ:\(x\ge2\)
\(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\dfrac{x-2}{4}}=26\\ \Leftrightarrow4\sqrt{x-2}+3\sqrt{x-2}-\dfrac{\sqrt{x-2}}{2}=26\\ \Leftrightarrow8\sqrt{x-2}+6\sqrt{x-2}-\sqrt{x-2}=52\\ \Leftrightarrow13\sqrt{x-2}=52\\ \Leftrightarrow\sqrt{x-2}=4\\ \Leftrightarrow x-2=16\\ \Leftrightarrow x=18\left(tm\right)\)
b,ĐKXĐ:\(x\in R\)
\(3x+\sqrt{4x^2-8x+4}=1\\ \Leftrightarrow2\sqrt{x^2-2x+1}=1-3x\\ \Leftrightarrow\left|x-1\right|=\dfrac{1-3x}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1-3x}{2}\\x-1=\dfrac{3x-1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-2=1-3x\\2x-2=3x-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
c, ĐKXĐ:\(x\ge0\)
\(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)-2\left(2\sqrt{x}+1\right)=7\\ \Leftrightarrow2x+\sqrt{x}-4\sqrt{x}-2=7\\ \Leftrightarrow2x-3\sqrt{x}-9=0\\ \Leftrightarrow\left(2x+3\sqrt{x}\right)-\left(6\sqrt{x}+9\right)=0\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+3\right)-3\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left(\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\2\sqrt{x}=-3\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=9\left(tm\right)\)
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a)
\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}=5-2x-x^2\)
\(\Leftrightarrow\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+16}=6-\left(x+1\right)^2\)
\(VT\ge6;VP\le6\Rightarrow VT=VP=6\)
Vậy pt có một nghiệm duy nhất là \(x=-1\)
b)
\(\sqrt{4x^2+20x+25}+\sqrt{x^2-8x+16}=\sqrt{x^2+18x+81}\)
\(\Leftrightarrow\sqrt{\left(2x+5\right)^2}+\sqrt{\left(x-4\right)^2}=\sqrt{\left(x+9\right)^2}\)
\(\Leftrightarrow\left|2x+5\right|+\left|x-4\right|=\left|x+9\right|\)
Lập bảng xét dấu ra nhé ~^o^~
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DKXD :\(x\ge-1\)
Đặt : \(\sqrt{x+1}=a\left(a\ge0\right)\Rightarrow\hept{\begin{cases}3x^2-8x-3=4xa\\a^2=x+1\end{cases}}\)
\(\Rightarrow3x^2-8x-3-4a^2=4xa-4a-4\Leftrightarrow4a^2+4xa+x^2=4x^2-4x+1\)
\(\Leftrightarrow\left(2a+x\right)^2=\left(2x-1\right)^2\)
+> \(2a+x=2x-1\Leftrightarrow2\sqrt{x+1}=x-1\Rightarrow4x+4=x^2-2x+1\left(x\ge1\right)\)
\(\Leftrightarrow x^2-6x-3=0\Rightarrow\orbr{\begin{cases}x=3+2\sqrt{3}\left(tm\right)\\3-2\sqrt{3}\left(ktm\right)\end{cases}}\)
+> \(2a+x=1-2x\Leftrightarrow2\sqrt{x+1}=1-3x\Rightarrow4x+4=9x^2-6x+1\left(x\le\frac{1}{3}\right)\)
\(\Leftrightarrow9x^2-10x-3=0\Rightarrow\orbr{\begin{cases}x=\frac{5+2\sqrt{13}}{9}\left(ktm\right)\\x=\frac{5-2\sqrt{13}}{9}\left(tm\right)\end{cases}}\)
Thử lại
Vậy :
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Đặt \(\sqrt{x^2+4x+7}=t>0\), ta có pt sau:
\(2\left(t^2+3\right)-7t=0\)
⇔ \(t^2-7t+6=0\Leftrightarrow\left(t-2\right)\left(2t-3\right)=0\)
⇔\(\left[{}\begin{matrix}t=2\\t=\frac{3}{2}\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x^2+4x+7=4\\x^2+4x+7=\frac{9}{4}\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\\x=\frac{\pm\sqrt{79}-4}{2}\end{matrix}\right.\)
Vậy ...
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Có cách này nhưng không chắc nha,mình mới lớp 7.
ĐK: \(8x^2-3x-1\ge0\) (để yên đi,đừng giải,xấu lắm)
\(4x^2+9x+1+\left(4x-1\right)\left(2x+1-\sqrt{8x^2-3x-1}\right)-\left(4x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow-4x^2+7x+2+\left(4x-1\right)\left[\frac{\left(2x+1\right)^2-8x^2+3x+1}{2x+1+\sqrt{8x^2-3x-1}}\right]=0\)
\(\Leftrightarrow-4x^2+7x+2+\left(4x-1\right)\left(\frac{-4x^2+7x+2}{2x+1+\sqrt{8x^2-3x-1}}\right)=0\)
\(\Leftrightarrow\left(-4x^2+7x+2\right)\left(1+\frac{4x-1}{2x+1+\sqrt{8x^2-3x-1}}\right)=0\)
Giải cái ngoặc nhỏ ta được hai nghiệm x = 2, x = -1/4 (t/m)
Giải tiếp cái ngoặc to thử xem sao (hình như vô nghiệm thì phải)
Thử giải tiếp cái ngoặc to nhé:
\(\frac{4x-1}{2x+1+\sqrt{8x^2-3x-1}}=-1\).Dễ thấy cái mẫu luôn khác 0 với mọi x.
Suy ra \(4x-1=-2x-1-\sqrt{8x^2-3x-1}\)
\(\Leftrightarrow6x=-\sqrt{8x^2-3x-1}\).Thêm đk: \(x\le0\),bình phương hai vế:
\(PT\Leftrightarrow28x^2+3x+1=0\).Phương trình này vô nghiệm.
Ta tìm được hai nghiệm: x = 2 hoặc x = -1/4
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Đk: \(x\ge-2\)
PT \(\Leftrightarrow\) \(x\left(12-4\sqrt{x+2}\right)+3x^2-20x-7=0\)
\(\Leftrightarrow x.\dfrac{144-16\left(x+2\right)}{12+4\sqrt{x+2}}+\left(x-7\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\dfrac{-4x\left(x-7\right)}{3+\sqrt{x+2}}+\left(x-7\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\\left(3+\sqrt{x+2}\right)\left(3x+1\right)=4x\end{matrix}\right.\)
Đặt \(u=\sqrt{x+2}\Leftrightarrow x=u^2-2\left(u\ge0\right)\)
PT (2) \(\Leftrightarrow\left(3+u\right)\left(3u^2-5\right)=4\left(u^2-2\right)\)
\(\Leftrightarrow9u^2-15+3u^3-5u=4u^2-8\)
\(\Leftrightarrow3u^3+5u^2-5u-7=0\) \(\Leftrightarrow u=\dfrac{-1+\sqrt{22}}{3}\)
\(\Leftrightarrow x=\dfrac{5-2\sqrt{22}}{9}\)
Vậy...
Lời giải:
ĐKXĐ: $x\geq -2$
PT $\Leftrightarrow 3x^2-20x-7=4x\sqrt{x+2}-12x$
$\Leftrightarrow (x-7)(3x+1)=4x(\sqrt{x+2}-3)=4x.\frac{x-7}{\sqrt{x+2}+3}$
$\Leftrightarrow x-7=0$ hoặc $3x+1=\frac{4x}{\sqrt{x+2}+3}$
Nếu $x-7=0\Leftrightarrow x=7$ (tm)
Nếu $3x+1=\frac{4x}{\sqrt{x+2}+3}$
$\Leftrightarrow 9x+3+(3x+1)\sqrt{x+2}=4x$
$\Leftrightarrow 5x+3+(3x+1)\sqrt{x+2}=0$
$\Leftrightaqrrow 5x+3=-(3x+1)\sqrt{x+2}$
$\Rightarrow (5x+3)^2=(3x+1)^2(x+2)$
$\Leftrightarrow 9x^3-x^2-17x-7=0$
$\Leftrightarrow (x+1)(9x^2-10x-7)=0$
$\Rightarrow$........