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a) \(2\left(x-5\right)-3\left(x+7\right)=14\)
\(\Leftrightarrow2x-10-3x-21=14\)
\(\Leftrightarrow-x-31=14\)
\(\Leftrightarrow-x=45\Leftrightarrow x=-45\)
b) \(5\left(x-6\right)-2\left(x+3\right)=12\)
\(\Leftrightarrow5x-30-2x-6=12\)
\(\Leftrightarrow3x-36=12\)
\(\Leftrightarrow3x=48\Leftrightarrow x=16\)
c) \(3\left(x-4\right)-\left(8-x\right)=12\)
\(\Leftrightarrow3x-12-8+x=12\)
\(\Leftrightarrow4x-20=12\)
\(\Leftrightarrow4x=32\Leftrightarrow x=8\)
d) \(-7\left(3x-5\right)+2\left(7x-14\right)=28\)
\(\Leftrightarrow-21x+35+14x-28=28\)
\(\Leftrightarrow-7x+35=0\Leftrightarrow x=5\)
e, \(18x-19=21+8x\)
\(\Rightarrow18x-8x=21+19\)
\(\Rightarrow10x=30\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
g, \(18-4x=-20-6x\)
\(\Rightarrow-4x+6x=-20-18\)
\(\Rightarrow-2x=-38\)
\(\Rightarrow x=19\)
Vậy \(x=19\)
h, \(-15x-24=-7x-32\)
\(\Rightarrow-15x+7x=-32+24\)
\(\Rightarrow-8x=-8\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
i, \(15x-3\left(4x-6\right)=-12+36\)
\(\Rightarrow15x-12x+18=-12+36\)
\(\Rightarrow7x=-12+36-18\)
\(\Rightarrow7x=6\)
\(\Rightarrow x=\frac{6}{7}\)
Vậy \(x=\frac{6}{7}\)
k, \(-10x-27=-7x+33\)
\(\Rightarrow-10x+7x=33+27\)
\(\Rightarrow-3x=60\)
\(\Rightarrow x=-20\)
m, \(-17x-24=-9x-40\)
\(\Rightarrow-17x+9x=-40+24\)
\(\Rightarrow-8x=-16\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
n, \(-23x-25=-18x+75\)
\(\Rightarrow-23x+18x=75+25\)
\(\Rightarrow-5x=100\)
\(\Rightarrow x=-20\)
Vậy \(x=-20\)
p, \(-5x+7\left(2x-3\right)=4\left(x-4\right)\)
\(\Rightarrow-5x+14x-21=4x-16\)
\(\Rightarrow-5x+14x-4x=-16+21\)
\(\Rightarrow5x=5\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
q, \(24-6\left(3x+1\right)=-5\left(4x-4\right)-8\)
\(\Rightarrow24-18x-6=-20x+20-8\)
\(\Rightarrow-18x+20x=20-8-24+6\)
\(\Rightarrow2x=-6\)
\(\Rightarrow x=-3\)
Vậy \(x=-3\)
r, \(18-4\left(6-2x\right)=-3\left(4x+5\right)-11\)
\(\Rightarrow18-24+8x=-12x-15-11\)
\(\Rightarrow8x+12x=-15-11-18+24\)
\(\Rightarrow20x=-20\)
\(\Rightarrow x=-1\)
Vậy \(x=-1\)
s, \(-5\left(4x-5\right)-14=6\left(2-3x\right)-10x-9\)
\(\Rightarrow-20x+25-14=12-18x-10x-9\)
\(\Rightarrow-20x+18x+10x=12-9+14-25\)
\(\Rightarrow8x=-8\)
\(\Rightarrow x=-1\)
Vậy \(x=-1\)
Chúc bạn hok tốt!!! tỷ tỷ
\(8-12x+6x^2-x^3\)
\(=\left(2-x\right)^3\)
\(125x^3-75x^2+15x-1\)
\(=\left(5x-1\right)^3\)
\(x^2-xz-9y^2+3yz\)
\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-z\right)\)
\(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^3+2x^2-6x-27\)
\(=x^3+5x^2+9x-3x^2-15x-27\)
\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
\(12x^3+4x^2-27x-9\)
\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(4x^2-9\right)\)
\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)
\(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)
5x.(-x)2+1=6
5x.x+1=6
5x2+1=6
5x2=6-1
5x2=5
x2=5:5
x2=1
x2=12
=>x=1
(15-x)+(x-12)=7-(-8+x)
15-x+x-12=7+8-x
3-x+x=15-x
3=15-x
x=15-3
x=12
4x3=4x
Để 4x3=4x=>x3=x
=>x=1
Câu cuối cùng mình ko bik.
hoc tốt
Ta có : \(\left|2x+4\right|+\left|4x+8\right|=0\left|2x+4\right|+\left|4x+8\right|=0\)
\(\Rightarrow\left|2x+4\right|+2.\left|2x+4\right|=\left|4x+8\right|=0\)
\(\Rightarrow\left|2x+4\right|\left(1+2\right)=0\)
=> |2x + 4| = 0
=> 2x + 4 = 0
=> 2x = -4
=> x = -2
1. Đề đúng phải là thế này: \(\left|2x+4\right|+\left|4x+8\right|=0\)
\(\Rightarrow\left|2x+4\right|=\left|4x+8\right|=0\)
\(\Rightarrow2x+4=4x+8=0\)
\(\Rightarrow x=-\frac{4}{2}=-\frac{8}{4}\)
\(\Rightarrow x=-2\)
2. Sửa lại đề : \(\left|x-5\right|-\left|x-7\right|=0\)
\(\Rightarrow\left|x-5\right|=\left|x-7\right|\)
\(\Rightarrow\orbr{\begin{cases}x-5=x-7\\x-5=-\left(x-7\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-5=-7\\x-5=-x+7\end{cases}}\)
( Loại trường hợp 1)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\)
3. \(\left|x+8\right|-\left|2x+2\right|=0\)
\(\Rightarrow\left|x+8\right|=\left|2x+2\right|\)
\(\Rightarrow\orbr{\begin{cases}x+8=2x+2\\x+8=-\left(2x+2\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x+2=8\\x+8=-2x-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=6\\3x=-10\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=6\\x=-\frac{10}{3}\end{cases}}\)