Đề trước đó: 

(x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x

<=>x^2+x-7x-7-x^2+6x-9=9x^2-25-9x^2-6x-1+x^2-4x+4-x

<=>x^2-11x-6=0

<=>x^2-2x. 11/2 + 121/4-145/4=0

<=>(x-11/2)^2=145/4

<=>|x-11/2|=căn(145)/2

<=>x=[11+-căn(145)]/2

cj ơi lỗi latex

a) \(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)

\(\Rightarrow\left(\frac{3}{5}-\frac{2}{3}-1\right).x=\frac{-5}{21}:\frac{1}{7}=\frac{-5}{3}\)

\(\Rightarrow\frac{-16}{15}.x=\frac{-5}{3}\Rightarrow x=\frac{-5}{3}:\frac{-16}{15}=\frac{25}{16}\)

b) \(\left(x-\frac{1}{4}\right)^2=\frac{1}{36}\)

\(\Rightarrow\left(x-\frac{1}{4}\right)^2=\left(±\frac{1}{6}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{4}=\frac{1}{6}\\x-\frac{1}{4}=\frac{-1}{6}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{12}\\x=\frac{1}{12}\end{cases}}\)

a: \(\left(2x^2+5x-2\right)\left(2x^2-4x+3\right)\)

\(=4x^4-8x^3+6x^2+10x^3-20x^2+15x-4x^2+8x-6\)

\(=4x^4+2x^3-18x^2+23x-6\)

b: \(\left(2x-3\right)\left(3x-2\right)-3x\left(2x-5\right)\)

\(=6x^2-4x-9x+6-6x^2+15x\)

=2x+6

c: \(\left(x-1\right)\left(x^2+x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)

\(=x^3-1-\left(x^3+1\right)\)

\(=x^3-1-x^3-1=-2\)

d: \(\left(x^2+x-1\right)\cdot\left(x^2-x+1\right)=\left(x^2\right)^2-\left(x-1\right)^2\)

\(=x^4-\left(x^2-2x+1\right)=x^4-x^2+2x-1\)

e: Sửa đề: \(\left(2x+3y\right)^2-\left(2x-3y\right)^2-12xy\)

=(2x+3y+2x-3y)(2x+3y-2x+3y)-12xy

=4x*6y-12xy

=24xy-12xy

=12xy

f: \(\left(x^2-4x\right)\left(5+2x-x^2\right)\)

\(=5x^2+2x^3-x^4-20x-8x^2+4x^3=-x^4+6x^3-3x^2-20x\)

(x-5)²=(1-3x)²

=> x-5 = 1- 3x

=> 4x = 6

=> x = \(\frac{3}{2}\)

( x - 5 )² = ( 1 - 3x ) ²

x - 5 = 1 - 3x

x = 1 - 3x + 5

x = 6 - 3x

x + 3x = 6

( 3 + 1 )x = 6

4x=6

=> x = 6 : 4

=> x = 1,5

\(\Leftrightarrow2x\left(x+5\right)-3\left(x-2\right)=7x+1\)

\(\Leftrightarrow2x^2+10x-3x+6-7x-1=0\)

\(\Leftrightarrow2x^2+5=0\)(vô lý)

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne2\\x\ne-5\end{matrix}\right.\)

\(\dfrac{2x}{x-2}-\dfrac{3}{x+5}=\dfrac{7x+1}{x^2+3x-10}\\ \Leftrightarrow\dfrac{2x\left(x+5\right)}{\left(x+5\right)\left(x-2\right)}-\dfrac{3\left(x-2\right)}{\left(x+5\right)\left(x-2\right)}=\dfrac{7x+1}{x^2-2x+5x-10}\\ \Leftrightarrow\dfrac{2x^2+10x}{\left(x+5\right)\left(x-2\right)}-\dfrac{3x-6}{\left(x+5\right)\left(x-2\right)}=\dfrac{7x+1}{x\left(x-2\right)+5\left(x-2\right)}\\ \Leftrightarrow\dfrac{2x^2+10x}{\left(x+5\right)\left(x-2\right)}-\dfrac{3x-6}{\left(x+5\right)\left(x-2\right)}-\dfrac{7x+1}{\left(x+5\right)\left(x-2\right)}=0\)

\(\Leftrightarrow\dfrac{2x^2+10x-3x+6-7x-1}{\left(x+5\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{2x^2+5}{\left(x+5\right)\left(x-2\right)}=0\\ \Rightarrow2x^2+5=0\left(vô.lí\right)\)

Vậy pt vô nghiệm

ta co : 2^x+2^x.2=48

<=>2^x(1+2)=48

<=>2^x  .  3=48

<=>2^x=48 : 3=16

<=>2^x=2^4

=>x=3

ta có: ( 5^x+1) - 5^x = 100

<=>5^x . 5 - 5^x = 100

<=> 5^x(5-1) = 100

<=> 5^x . 4= 100

<=> 5^x= 100: 4=25

<=> 5^x= 5^2

=> x=2

1, Ta có :

     \(x+\frac{3}{5}=\frac{4}{7}\div\frac{8}{21}\)

    \(x+\frac{3}{5}=\frac{4}{7}\times\frac{21}{8}\)

    \(x+\frac{3}{5}=\frac{3}{2}\)

    \(x=\frac{3}{2}-\frac{3}{5}\)

    \(x=\frac{15}{10}-\frac{6}{10}\)

    \(x=\frac{9}{10}\)

Vậy x = \(\frac{9}{10}\)

2, Ta có :

    \(\frac{2}{3}+\frac{3}{4}\div x=-\frac{1}{6}\)

    \(\frac{3}{4}\div x=-\frac{1}{6}-\frac{2}{3}\)

    \(\frac{3}{4}\div x=-\frac{1}{6}-\frac{4}{6}\)

    \(\frac{3}{4}\div x=-\frac{5}{6}\)

    \(x=\frac{3}{4}\div\left(-\frac{5}{6}\right)\)

    \(x=\frac{3}{4}\times\left(-\frac{6}{5}\right)\)

    \(x=-\frac{9}{10}\)

Vậy x = \(-\frac{9}{10}\)

thanks bạn nha

x²-3x+2=6

=>x²-3x=4

=>x.(x-3)=4

=>x và x-3 thuộc Ư(4)

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