x^8 -25x^6 =0
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![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\left(x^2+x\right)^2+2\left(x^2+x\right)-8=0\)
\(\Leftrightarrow\left(x^2+x+4\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
hay \(x\in\left\{-2;1\right\}\)
b: \(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x+2\right)\left(x+4\right)+24=0\)
\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x-12\right)+24=0\)
\(\Leftrightarrow\left(x^2+x\right)^2-14\left(x^2+x\right)+48=0\)
\(\Leftrightarrow\left(x^2+x-6\right)\left(x^2+x-8\right)=0\)
hay \(x\in\left\{-3;2;\dfrac{-1+\sqrt{33}}{2};\dfrac{-1-\sqrt{33}}{2}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
y: Ta có: \(x^2-x-6=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
z: Ta có: \(3x^2-5x-8=0\)
\(\Leftrightarrow\left(3x-8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-1\end{matrix}\right.\)
j: Ta có: \(25x^2-4=0\)
\(\Leftrightarrow\left(5x-2\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
6) \(\sqrt{x^2-4x+1}=x\left(x\ge0\right)\)
\(\Leftrightarrow x^2-4x+1=x^2\)
\(\Leftrightarrow x^2-x^2=4x-1\)
\(\Leftrightarrow4x=1\)
\(\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)
8) \(\sqrt{x^2-x-6}=\sqrt{x-3}\left(x\ge3\right)\)
\(\Leftrightarrow x^2-x-6=x-3\)
\(\Leftrightarrow x^2-2x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
9) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\left(x\ge1\right)\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=1+1\)
\(\Leftrightarrow x=2\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A) x3-6x2+12x-8=0
<=>(x-2)3=0
<=>x-2=0
<=>x=2
B)4(x-3)2 -(2x-1)(2x+1)=13
<=>4(x2-6x+9)-4x2+1=13
<=>4x2-24x+36-4x2+1=13
<=>-24x+37=13
<=>24x=37-13
<=>24x=24
<=>x=1
C)25x2-6(x+1)2=0
<=>(5x-\(\sqrt{6}\left(x+1\right)\))(5x+\(\sqrt{6}\left(x+1\right)\))=0
<=>5x-\(\sqrt{6}\left(x+1\right)\)=0 hoặc 5x+\(\sqrt{6}\left(x+1\right)\))=0
<=>5x-\(\sqrt{6}x-\sqrt{6}\)=0 <=>5x+\(\sqrt{6}x+\sqrt{6}\)=0
<=>x(5-\(\sqrt{6}\))=\(\sqrt{6}\) <=>x(5+\(\sqrt{6}\))=\(-\sqrt{6}\)
<=>x=\(\frac{\sqrt{6}}{5-\sqrt{6}}\) <=>x=\(\frac{-\sqrt{6}}{5+\sqrt{6}}\)
Rút gọn C=(4+2A+A^2).(4-A^2).(4-2a+a^2) GIẢI GIÚP MIK ĐI
![](https://rs.olm.vn/images/avt/0.png?1311)
24+50+25x=82+62
24+1+25x=64+36
25+25x=100
25x=100-25
25x=75
x=75:25
x=3
vậy x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
1) 2x4 - 9x3 + 14x2 - 9x + 2 = 0
<=> (2x4 - 4x3) - (5x3 - 10x2) + (4x2 - 8x) - (x - 2) = 0
<=> 2x3(x - 2) - 5x2(x - 2) + 4x(x - 2) - (x - 2) = 0
<=> (2x3 - 5x2 + 4x - 1)(x - 2) = 0
<=> [(2x3 - 2x2) - (3x2 - 3x) + (x - 1)](x - 2) = 0
<=> [2x2(x - 1) - 3x(x - 1) + (x - 1)](x - 2) = 0
<=> (2x2 - 2x - x + 1)(x - 1)(x - 2) = 0
<=> (2x - 1)(x - 1)2(x - 2) = 0
<=> 2x - 1=0
hoặc x - 1 = 0
hoặc x - 2 = 0
<=> x = 1/2
hoặc x = 1
hoặc x = 2
Vậy S = {1/2; 1; 2}
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: x>=-1
Sửa đề: \(6\sqrt{x+1}-\sqrt{25x+25}+8\sqrt{\dfrac{x+1}{4}}=10\)
=>\(6\sqrt{x+1}-5\sqrt{x+1}+8\cdot\dfrac{\sqrt{x+1}}{2}=10\)
=>\(\sqrt{x+1}+4\sqrt{x+1}=10\)
=>\(5\sqrt{x+1}=10\)
=>\(\sqrt{x+1}=2\)
=>x+1=4
=>x=3(nhận)
bạn là ai thế