Tìm x:
1)3x^2-17x+10=0
2)x^4-4x^3+5x^2-4x+4+0
K
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2
1 tháng 3 2022
a: \(\Leftrightarrow4x^2+9x-4x-9=0\)
=>(4x+9)(x-1)=0
=>x=1 hoặc x=-9/4
b: \(\Leftrightarrow x^2-x-4x+4=0\)
=>(x-1)(x-4)=0
=>x=1 hoặc x=4
c: \(\Leftrightarrow5x^2-5x-12x+12=0\)
=>(x-1)(5x-12)=0
=>x=12/5 hoặc x=1
d: \(\Leftrightarrow x^2-4x+x-4=0\)
=>(x-4)(x+1)=0
=>x=4 hoặc x=-1
1 tháng 3 2022
a, Ta có a + b + c = 4 + 5 - 9 = 0
vậy pt có 2 nghiệm x = 1 ; x = -9/4
b, Ta có a + b + c = 1 - 5 + 4 = 0
vậy pt có 2 nghiệm x = 1 ; x = 4
c, Ta có a + b + c = 5 - 17 + 12 = 0
vậy pt có 2 nghiệm x = 1 ; x = 12/5
d, Ta có a - b + c = 1 + 3 - 4 = 0
vậy pt có 2 nghiệm x = -1 ; x = 4
NT
2
TD
9 tháng 2 2018
1.\(\left|9-7x\right|=5x-3\)
\(\Leftrightarrow\orbr{\begin{cases}9-7x=5x-3\\9-7x=-5x-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-7x-5x=-9-3\\-7x+5x=-9-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-12x=-12\\-2x=-12\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-12:\left(-12\right)\\x=-12:\left(-2\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=6\end{cases}}\)
2.\(8x-\left|4x+1\right|=x+2\)
\(\Rightarrow\left|4x+1\right|=8x-x+2\)
\(\Rightarrow\left|4x+1\right|=7x+2\)
\(\Leftrightarrow\orbr{\begin{cases}4x+1=7x+2\\4x+1=-7x+2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x-7x=2-1\\4x+7x=2-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-3x=1\\11x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1:\left(-3\right)\\x=1:11\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=\frac{1}{11}\end{cases}}\)
KT
1
3 tháng 8 2020
1) Ta có: \(\left(x^2-4x+4\right)\left(x^2+4x+4\right)-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)^2\cdot\left(x+2\right)^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\right]^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x^2-4\right)^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x^2-4-7x-4\right)\left(x^2-4+7x+4\right)=0\)
\(\Leftrightarrow\left(x^2-7x-8\right)\left(x^2+7x\right)=0\)
\(\Leftrightarrow x\left(x+7\right)\left(x^2-8x+x-8\right)=0\)
\(\Leftrightarrow x\left(x+7\right)\left[x\left(x-8\right)+\left(x-8\right)\right]=0\)
\(\Leftrightarrow x\left(x+7\right)\left(x-8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+7=0\\x-8=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\\x=8\\x=-1\end{matrix}\right.\)
Vậy: S={0;-7;8;-1}
2) Ta có: \(x^3-8x^2+17x-10=0\)
\(\Leftrightarrow x^3-2x^2-6x^2+12x+5x-10=0\)
\(\Leftrightarrow x^2\left(x-2\right)-6x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-6x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-x-5x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=5\end{matrix}\right.\)
Vậy: S={2;1;5}
3) Ta có: \(2x^3-5x^2-x+6=0\)
\(\Leftrightarrow2x^3-4x^2-x^2+2x-3x+6=0\)
\(\Leftrightarrow2x^2\left(x-2\right)-x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-3x+2x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(2x-3\right)+\left(2x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\2x=3\\x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{3}{2}\\x=-1\end{matrix}\right.\)
Vậy: \(S=\left\{2;\frac{3}{2};-1\right\}\)
4) Ta có: \(4x^4-4x^2-3=0\)
\(\Leftrightarrow4x^4-6x^2+2x^2-3=0\)
\(\Leftrightarrow2x^2\left(2x^2-3\right)+\left(2x^2-3\right)=0\)
\(\Leftrightarrow\left(2x^2-3\right)\left(2x^2+1\right)=0\)
mà \(2x^2+1>0\forall x\in R\)
nên \(2x^2-3=0\)
\(\Leftrightarrow2x^2=3\)
\(\Leftrightarrow x^2=\frac{3}{2}\)
hay \(x=\pm\sqrt{\frac{3}{2}}\)
Vậy: \(S=\left\{\sqrt{\frac{3}{2}};-\sqrt{\frac{3}{2}}\right\}\)