a)2x.(3x+5)-x.(6x-1)=33

=>\(6x^2+10x-6x^2+x=33\)

=>11x=33

=>x=3

b)x(3x-1)+12x-4=0

=>x(3x-1)+4(3x-1)=0

=>(x-4)(3x-1)=0

=>x-4=0 hoặc 3x-1=0

+)x-4=0 +)3x-1=0

=>x=4 =>x=\(\frac{1}{3}\)

c)5x(2x+1)-12x-6=0

=>10x\(^2\)+5x-12x-6=0

=>10x\(^2\)-7x-6=0

=>(10x\(^2\)+5x)-(12x+6)=0

=>5x(2x+1)-6(2x+1)=0

=>(5x-6)(2x+1)=0

=>\(\left[{}\begin{matrix}5x-6=0\\2x+1=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\frac{6}{5}\\x=\frac{-1}{2}\end{matrix}\right.\)

a,

128-3x-12=23

3x=128-12-23

3x=93

x=93:3

= 31

b,

(12x+84+55):5=35

12x+84+55=35.5

12x+84+55=175

12x=175-55-84

12x=36

x=36:12

x=3

\(d,=24x^2-38x+3\\ e,=x^2-12x+35\\ f,=\left(x^2-144\right)\left(4x-1\right)=4x^3-x^2-576x+144\)

Tính giá trị biểu thức

ý d, e, f

d: \(\left(12x-1\right)\left(2x-3\right)\)

\(=24x^2-36x-2x+3\)

\(=24x^2-38x+3\)

\(a,128-3.\left(x+4\right)=23\\ \Rightarrow3.\left(x+4\right)=105\\ \Rightarrow x+4=35\\ \Rightarrow x=31\\ b,\left[\left(4x+28\right).3+55\right]:5=35\\ \Rightarrow\left(4x+28\right).3+55=175\\ \Rightarrow4x+28.3=120\\ \Rightarrow4x+28=60\\ \Rightarrow4x=32\\ \Rightarrow x=8.\)

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) \(720:\left[41-\left(2x-5\right)\right]:5=35\)

\(720:\left(41-2x+5\right):5=35\)

\(720:\left(46-2x\right)=35.5\)

\(720:\left(46-2x\right)=175\)

\(46-2x=720:175\)

\(46-2x=\dfrac{144}{35}\)

\(2x=46-\dfrac{144}{35}\)

\(2x=\dfrac{1466}{35}\)

\(x=\dfrac{1466}{35}:2\)

\(x=\dfrac{733}{35}\)

lm nốt hộ mik 2 ý cuối vs ạ

a)128-3(x+4)=23 

    125.(x+4)=23

             ( x +4)=23 : 125

             x+4 = 0,184

                x = 0,184 -4

                  x = -3 .816

theo mềnh nhóe ( làm từng câu )

c: =>12x-64=32

=>12x=96

hay x=8

g: =>3x-16=14

hay x=10

e: =>5(x+4)=85

=>x+4=17

hay x=13

1,=\(x^2-3x-2x^2+6x=-x^2+3x\)

2,=\(3x^2-x-5+15x=3x^2+14x-5\)

3,=\(5x+15-6x^2-6x=-6x^2-x+15\)

4,=\(4x^2+12x-x-3=4x^2+11x-3\)

5: =>(x+5)^3=0

=>x+5=0

=>x=-5

6: =>(2x-3)^2=0

=>2x-3=0

=>x=3/2

7: =>(x-6)(x-10)=0

=>x=10 hoặc x=6

8: \(\Leftrightarrow x^3-12x^2+48x-64=0\)

=>(x-4)^3=0

=>x-4=0

=>x=4

a, 128 – 3(x+4) = 23

b,  12 x - 4 3 . 8 3 = 4 . 8 4

c, [(4x+28).3+55]:5 = 35

d, 720:[41 – (2x – 5)] =  2 3 . 5

128-3.(x+4) =23                 

        3.(x+4) =128-23

        3.(x+4) = 105

            x+4 =105:3

            x+4 =35

               x = 35-4

               x=31

4.x + 5.x -x =128

a) \(x^2-12x+11\)\(=0\)

\(\Leftrightarrow\left(x-6\right)^2-25=0\)

\(\Leftrightarrow\left(x-6+5\right)\left(x-6-5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=11\end{matrix}\right.\)

a)\(x^2-12x+11=0\)

\(x^2-x-11x+11=0\)

\(\left(x^2-x\right)-\left(11x-11\right)=0\)

\(x\left(x-1\right)-11\left(x-1\right)=0\)

\(\left(x-1\right)\left(x-11\right)=0\)

\(=>\left[{}\begin{matrix}x-1=0\\x-11=0\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x=1\\x=11\end{matrix}\right.\)

b)\(4x^2-4x-3=0\)

\(4x^2-2x+6x-3=0\)

\(2x\left(2x-1\right)+3\left(3x-1\right)=0\)

\(\left(2x-1\right)\left(2x+3\right)=0\)

\(=>\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x=0,5\\x=-1,5\end{matrix}\right.\)\

c)\(4x^2-12x-7=0\)

\(4x^2-14x+2x-7=0\)

\(2x\left(2x-7\right)+\left(2x-7\right)=0\)

\(\left(2x-7\right)\left(2x+1\right)=0\)

\(=>\left[{}\begin{matrix}2x-7=0\\2x+1=0\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)

d)\(x^3-6x^2=8-12x\)

\(=>\left(x^3-6x^2\right)-\left(8-12x\right)=0\)

\(=>x^3-6x^2-8+12x=0\)

\(x^3-3x^2.2+3x.2^2-2^3=0\)

\(\left(x-2\right)^3=0\)

\(=>x-2=0\)

\(=>x=2\)