8.(x+2)-6.x=140
10.(x-4)-8.(x+2)=12
x.(x-3)=6.(x-3)
K
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26 tháng 10 2021
6) ĐKXĐ: \(x\le-6\)
\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)
\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)
Vậy \(x\le-6\)
7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)
\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)
\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)
Vậy \(x\ge\dfrac{2}{3}\)
8) ĐKXĐ: \(x\ge5\)
\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)
\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)
9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
DH
8 tháng 7 2017
Bài 4:
a, \(x^3+12x^2+48x+64=x^3+4x^2+8x^2+32x+16x+64\)
\(=x^2.\left(x+4\right)+8x.\left(x+4\right)+16.\left(x+4\right)\)
\(=\left(x+4\right).\left(x^2+8x+16\right)=\left(x+4\right).\left(x^2+4x+4x+16\right)\)
\(=\left(x+4\right).\left(x+4\right)^2=\left(x+4\right)^3\)(1)
Thay \(x=6\) vào (1) ta được:
\(\left(6+4\right)^3=10^3=1000\)
Vậy...........
b, \(x^3-6x^2+12x-8=x^3-2x^2-4x^2+8x+4x-8\)
\(=x^2.\left(x-2\right)-4x.\left(x-2\right)+4.\left(x-2\right)\)
\(=\left(x-2\right).\left(x^2-4x+4\right)=\left(x-2\right).\left(x^2-2x-2x+4\right)\)
\(=\left(x-2\right).\left(x-2\right)^2=\left(x-2\right)^3\)(2)
Thay \(x=22\) vào (2) ta được:
\(\left(22-2\right)^3=20^3=8000\)
Vậy.............
Chúc bạn học tốt!!!
DH
8 tháng 7 2017
Bài 2:
a, \(\left(x+9\right)^3=27=3^3\)
\(\Rightarrow x+9=3\Rightarrow x=-6\)
Vậy.........
b, \(8-12x-x^3+6x^2=-64\)
\(\Rightarrow-\left(x^3-6x^2+12x-8\right)=-64\)
\(\Rightarrow x^3-2x^2-4x^2+8x+4x-8=64\)
\(\Rightarrow x^2.\left(x-2\right)-4x.\left(x-2\right)+4.\left(x-2\right)=64\)
\(\Rightarrow\left(x-2\right).\left(x^2-4x+4\right)=64\)
\(\Rightarrow\left(x-2\right).\left(x^2-2x-2x+4\right)=64\)
\(\Rightarrow\left(x-2\right).\left(x-2\right)^2=64\)
\(\Rightarrow\left(x-2\right)^3=4^3\Rightarrow x-2=4\Rightarrow x=6\)
Vậy............
Chúc bạn học tốt!!!
AD
23 tháng 7 2023
\(a,\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\\ b,=\left(2x\right)^2+2.2x.3+3^2\\ =\left(2x+3\right)^2\\ c,=x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ d,=x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)
\(e,=-4x^2\left(x-1\right)+\left(x-1\right)\\ =\left(1-4x^2\right)\left(x-1\right)\\ =\left(1-2x\right)\left(1+2x\right)\left(x-1\right)\)
\(f,=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\\ =\left(2x+1\right)^3\)
LV
1
25 tháng 7 2023
a: \(=\dfrac{3\left(x-2\right)}{\left(x-2\right)^3}=\dfrac{3}{\left(x-2\right)^2}\)
b: \(=\dfrac{x^2\left(x+2\right)}{\left(x+2\right)^3}=\dfrac{x^2}{\left(x+2\right)^2}\)
H9
HT.Phong (9A5)
CTVHS
31 tháng 7 2023
1) \(8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x\right)^2-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
2) \(x^3-6x^2+12x-8=27\)
\(\Leftrightarrow x^3-3\cdot x^2\cdot2+3\cdot2^2\cdot x-2^3=27\)
\(\Leftrightarrow\left(x-2\right)^3=27\)
\(\Leftrightarrow\left(x-2\right)^3=3^3\)
\(\Leftrightarrow x-2=3\)
\(\Leftrightarrow x=3+2\)
\(\Leftrightarrow x=5\)
3) \(x^2-8x+16=5\left(4-x\right)^3\)
\(\Leftrightarrow\left(x-4\right)^2=5\left(4-x\right)^3\)
\(\Leftrightarrow\left(4-x\right)^2=5\left(4-x\right)^3\)
\(\Leftrightarrow5\left(4-x\right)=1\)
\(\Leftrightarrow4-x=\dfrac{1}{5}\)
\(\Leftrightarrow x=4-\dfrac{1}{5}\)
\(\Leftrightarrow x=\dfrac{19}{5}\)
4) \(\left(2-x\right)^3=6x\left(x-2\right)\)
\(\Leftrightarrow8-12x+6x^2-x^3=6x^2-12x\)
\(\Leftrightarrow-12x+6x^2-6x^2+12x=8-x^3\)
\(\Leftrightarrow8-x^3=0\)
\(\Leftrightarrow x^3=8\)
\(\Leftrightarrow x^3=2^3\)
\(\Leftrightarrow x=2\)
5) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x-3x\right)+\left(3x^2+3x^2\right)+\left(1+1\right)-6x^2+12x-6=-10\)
\(\Leftrightarrow0+0+0+\left(6x^2-6x^2\right)+12x-4=-10\)
\(\Leftrightarrow12x-4=-10\)
\(\Leftrightarrow12x=-10+4\)
\(\Leftrightarrow12x=-6\)
\(\Leftrightarrow x=\dfrac{-6}{12}\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
6) \(\left(3-x\right)^3-\left(x+3\right)^3=36x^2-54x\)
\(\Leftrightarrow27-27x+9x^2-x^3-x^3-9x^2-27x-27=36x^2-54x\)
\(\Leftrightarrow-54x-2x^3=36x^2-54x\)
\(\Leftrightarrow-2x^3=36x^2\)
\(\Leftrightarrow-2x^3-36x^2=0\)
\(\Leftrightarrow-2x^2\left(x+18\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x^2=0\\x+18=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-18\end{matrix}\right.\)
GL
24 tháng 8 2023
\(x^2+2x-8\)
\(=x^2+4x-2x-8\)
\(=x^2\left(x+4\right)-2\left(x+4\right)\)
\(=\left(x^2-2\right)\left(x+4\right)\)
\(x^2+5x+6\)
\(=x^2+2x+3x+6\)
\(=x\left(x+2\right)+3\left(x+2\right)\)
\(=\left(x+3\right)\left(x+2\right)\)
\(4x^2-12x+8\)
\(=4x^2-4x-8x+8\)
\(=4x\left(x-1\right)-8\left(x-1\right)\)
\(=\left(4x-8\right)\left(x-1\right)\)
\(x^2-xy-\dfrac{3}{4}y^2\)
\(=x^2-\dfrac{3}{2}xy+\dfrac{1}{2}xy-\dfrac{3}{4}y^2\)
\(=x\left(x-\dfrac{3}{2}y\right)+\dfrac{1}{2}y\left(x-\dfrac{3}{2}y\right)\)
\(=\left(x+\dfrac{1}{2}y\right)\left(x-\dfrac{3}{2}y\right)\)
4 tháng 10 2021
1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)
2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)
4) \(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)
6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)
7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)
8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)
9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)
10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)
11) \(=\left(x+2\right)^3\)
12) \(=\left(x+3\right)^3\)
18 tháng 6 2022
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 \(x\in\left\{-3;2;\dfrac{-1+\sqrt{33}}{2};\dfrac{-1-\sqrt{33}}{2}\right\}\)
KD
0
10( x - 4 ) - 8( x + 2 ) = 12
10x - 40 - 8x + 16 = 12
=> x( 10 - 8 ) - 40 + 16 = 12
2x - 40 + 16 = 12
2x - 40 = -4
2x = -4 + 40 = 36
x = 36 : 2 = 18
8.(x+2)-6.x=140
8( x + 2 ) - 6x = 140
8x + 16 - 6x = 140
x( 8 - 6 ) + 16 = 140
2x + 16 = 140
2x = 140 - 16
2x = 124
x = 124 : 2
x= 62
10.(x-4)-8.(x+2)=12
10x - 40 - 8x + 16 = 12
x( 10 - 8 ) - 40 + 16 = 12
2x - 40 + 16 = 12
2x - 40 = 12-16
2x - 40 = -4
2x = -4 + 40
2x = 36
x = 36 : 2
x= 18
x.(x-3)=6.(x-3)