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a, 7x + 10x = 5x
17x = 5x
17x - 5x = 0
12x = 0
x =0
2;
a, 4x + 7x = 22
11x = 22
x = 2
b, 12x - 8x = 25
4x = 25
x = \(\dfrac{25}{4}\)
c, \(\dfrac{1}{2}\)x - \(\dfrac{1}{3}\)x = \(\dfrac{4}{5}\)
(\(\dfrac{1}{2}-\dfrac{1}{3}\))x = \(\dfrac{4}{5}\)
\(\dfrac{1}{6}\)x = \(\dfrac{4}{5}\)
x = \(\dfrac{4}{5}\) : \(\dfrac{1}{6}\)
x = \(\dfrac{24}{5}\)
52-2(3x-6)=10 3636:(12x-91)=36 2x+3=32
2(3x-6)=52-10 12x-91=3636:36 2x=32-3
2(3x-6)=42 12x-91=101 2x=29
3x-6=42:2 12x=101+91 x=29:2 (loại vì x thuộc N)
3x-6=21 12x=192
3x=21+6 x=192:12=16
3x=27
x=27:3=9
2x.8=512 (24-x)3=64
2x=512:8 (24-x)3=43
2x=64 => 24-x=4
x=64:2=32 x=24-4=20
a,52-2(3x-6)=10
2(3x-6)=52-19=42
3x-6=42:2=21
3x=21+6=27
x=27:3
x=9
b,3636:(12x-91)=36
12x-91=3636:36=101
12x=101+91=192
x=192:12
x=16
c,2x+3=32
2x=32-3=29
x=29:2
x=29/2.Vì x thuộc tập số tự nhiên=> không tồn tại x thỏa mãn đề bài
d,2x.8=512
2x=512:8=64
x=64:2
x=32
e,(24-x)^3=64
(24-x)^3=4^3
=>24-x=4
x=24-4
x=20
\(1,\)
\(A=-\frac{7}{12}+\frac{12}{18}+\frac{5}{4}\)
\(=-\frac{7}{12}+\frac{2}{3}+\frac{5}{4}\)
\(=-\frac{7}{12}+\frac{8}{12}+\frac{15}{12}\)
\(=\frac{-7+8+15}{12}\)
\(=\frac{4}{3}\)
\(1,\)
\(B=\frac{1}{4}-\frac{8}{7}:8-3:\frac{3}{4}.\left(-2\right)^2\)
\(=\frac{1}{4}-\frac{8}{7}.\frac{1}{8}-3.\frac{4}{3}.4\)
\(=\frac{1}{4}-\frac{1}{7}-16\)
\(=\frac{7-4-448}{28}\)
\(=-\frac{445}{28}\)
Mình đang bận, bạn cần gấp thế à? Trả lời trong tin nhắn nhé!!!
\(8-12x+6x^2-x^3\)
\(=\left(2-x\right)^3\)
\(125x^3-75x^2+15x-1\)
\(=\left(5x-1\right)^3\)
\(x^2-xz-9y^2+3yz\)
\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-z\right)\)
\(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^3+2x^2-6x-27\)
\(=x^3+5x^2+9x-3x^2-15x-27\)
\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
\(12x^3+4x^2-27x-9\)
\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(4x^2-9\right)\)
\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)
\(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)
a: =>3(x+4)=115
=>x+4=115/3
hay x=103/3
b: =>\(8^3\cdot\left(12x-64\right)=4\cdot8^4\)
\(\Leftrightarrow12x-64=32\)
=>12x=96
hay x=8
c: \(\Leftrightarrow\left(4x+28\right)\cdot23+55=175\)
=>(4x+28)x23=120
=>4x+28=120/23
=>4x=-524/23
hay x=-131/23
\(a,128-3.\left(x+4\right)=23\\ \Rightarrow3.\left(x+4\right)=105\\ \Rightarrow x+4=35\\ \Rightarrow x=31\\ b,\left[\left(4x+28\right).3+55\right]:5=35\\ \Rightarrow\left(4x+28\right).3+55=175\\ \Rightarrow4x+28.3=120\\ \Rightarrow4x+28=60\\ \Rightarrow4x=32\\ \Rightarrow x=8.\)
c) \(\left(12x-4^3\right).8^3=4.8^4\)
\(12x-64=4.8^4:8^3\)
\(12x-64=32\)
\(12x=32+64\)
\(12x=96\)
\(x=\dfrac{96}{12}\)
\(x=8\)
d) \(720:\left[41-\left(2x-5\right)\right]:5=35\)
\(720:\left(41-2x+5\right):5=35\)
\(720:\left(46-2x\right)=35.5\)
\(720:\left(46-2x\right)=175\)
\(46-2x=720:175\)
\(46-2x=\dfrac{144}{35}\)
\(2x=46-\dfrac{144}{35}\)
\(2x=\dfrac{1466}{35}\)
\(x=\dfrac{1466}{35}:2\)
\(x=\dfrac{733}{35}\)
\(a,-4\left(2x+9\right)=\left(-8x+3\right)\)
\(\Rightarrow-8x-36=-8x+3\)
\(\Rightarrow-8x+8x=3+36\)
\(\Rightarrow0x=39\left(vô-lí\right)\)
\(b,1+x-2\left(5+3x\right)=4-5x\)
\(\Rightarrow1+x-10-6x=4-5x\)
\(\Rightarrow x-6x+5x=4+10-1\)
\(\Rightarrow0x=13\left(vô-lí\right)\)
\(c,3\left(2-x\right)+1=-3x+7\)
\(\Rightarrow6-3x+1=-3x+7\)
\(\Rightarrow-3x+3x=7-1-6\)
\(\Rightarrow0x=0\Rightarrow x=0\)
5(x + -7) + -4(x + -1) = 8(x + 3) + -12x
5(-7 + x) + -4(x + -1)
= 8(x + 3) + -12x (-7 * 5 + x * 5) + -4(x + -1)
= 8(x + 3) + -12x (-35 + 5x) + -4(x + -1)
= 8(x + 3) + -12x