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ĐKXĐ \(x^2-7x+8\ge0\)
\(\Rightarrow x^2-7x+8+\sqrt{x^2-7x+8}=20\)
Đặt a = \(\sqrt{x^2-7x+8}\) (a \(\ge\)0) ta đc:
\(a^2+a=20\)
\(\Rightarrow a^2+a-20=0\)
\(\Rightarrow a=4\) hoặc \(a=-5\) (loại)
Với a = 4
<=> \(\sqrt{x^2-7x+8}=4\)
\(\Leftrightarrow x^2-7x+8=16\)
\(\Leftrightarrow x^2-7x-8=0\)
\(\Rightarrow\left(x-8\right)\left(x+1\right)=0\)
=> x - 8 = 0 => x = 8
hoặc x + 1 = 0 => x = -1
Vậy x = 8 ; x = -1
\(\Leftrightarrow\left(x^2-9x+8\right)\left(x^2-6x+8\right)=7x^2\)
Xét x=0: x=0 không là nghiệm của phương trình
Xét x\(\ne\)0
pt \(\Leftrightarrow\left(x-9+\frac{8}{x}\right)\left(x-6+\frac{8}{x}\right)=7\)
Đặt t= x+8/x
Sau đó bạn giải pt tìm t, có t thế vào tìm được x
a)\(\left\{{}\begin{matrix}3x-2y=3\\2x+2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=5\\3x-2y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\3-2y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)
b)\(x^2+7x+12=0\)
\(\Leftrightarrow x^2+3x+4x+12=0\)( chị nghĩ + 12 đúng hơn á )
\(\Leftrightarrow x\left(x+3\right)+4\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=-4\end{matrix}\right.\)
a)x5+x-1=0
<=>(x5+x4+x3+x2+x)-(x4+x3+x2+x+1)=0
<=>(x4+x3+x2+x+1)(x-1)=0
Do x4+x3+x2+x+1>0
=>x+1=0
<=>x=1
đk: \(x\ge0\)
\(x\sqrt{x}+4\sqrt{x}+12=7x\)
\(\Leftrightarrow\left(x+4\right)\sqrt{x}=7x-12\)
\(\Leftrightarrow\left(x+4\right)^2\cdot x=\left(7x-12\right)^2\)
\(\Leftrightarrow x^3+8x^2+16x=49x^2-168x+144\)
\(\Leftrightarrow x^3-41x^2+184x-144=0\)
\(\Leftrightarrow\left(x^3-x^2\right)-\left(40x^2-40x\right)+\left(144x-144\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-40x+144\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-4\right)\left(x-36\right)=0\)
=> \(x\in\left\{1;4;36\right\}\)