\(a,8x-3=5x+12\\ \Leftrightarrow8x-5x=12+3\\ \Leftrightarrow3x=15\\ \Leftrightarrow x=\dfrac{15}{3}=5\)

\(b,x-12+4x=25+2x-1\\ \Leftrightarrow x+4x-2x=25-1+12\\ \Leftrightarrow3x=36\\ \Leftrightarrow x=\dfrac{36}{3}=12\)

\(c,7-\left(2x+4\right)=-\left(x+4\right)\\ \Leftrightarrow7-2x-4=-x-4\\ \Leftrightarrow-2x+x=-4+4-7\\ \Leftrightarrow-x=-7\\ \Leftrightarrow x=7\)

\(d,3-4x\left(45-2x\right)=8x^2+x-300\\ \Leftrightarrow3-100x+8x^2=8x^2+x-300\\ \Leftrightarrow8x^2-8x^2-100x-x=-300-3\\ \Leftrightarrow-101x=-303\\ \Leftrightarrow x=\dfrac{-303}{-101}=3\)

Đề câu d của bạn hình như sai dấu ý

sửa lại

.d. 3 - 4x (25 - 2x) = 8x ² + x - 300

\(3-4x\left(25-2x\right)=8x^2+x-300.\)

\(3-100x+8x^2=8x^2+x-300\)

\(3-100x=x-300\)

\(3+300=x+100x\)

\(303=101x\)

\(x=3\)

Vậy x cần tìm bằng 3

3-4x(25-2x)=8x^2 + x -300 
<=> 3-100x+8x^2=8x^2 + x -300 
<=>3-100x=x-300 
<=>101x=303 
<=>x=3 

\(3-4x\left(25-2x\right)=8x^2+x-300\)

\(\Leftrightarrow3-100x+8x^2=8x^2+x-300\)\(\Leftrightarrow8x^2-8x^2-100x-x=-3-300\)

\(\Leftrightarrow-101x=-303\)

\(\Leftrightarrow x=3\)

\(3-4x\left(25-2x\right)=8x^2+x-300\)

\(\Leftrightarrow3-100x+8x^2=8x^2+x-300\)

\(\Leftrightarrow-101x=-303\)

\(\Leftrightarrow x=3\)

Vậy phương trình có nghiệm \(x=3\)

Chúc bạn học tốt!

bạn gõ công thức toán đi ! như này khó nhìn quá :((

Bài làm

3 - 4x( 25 - 2x ) = 8x² - x - 300

<=> 3 - 100x + 8x² - 8x² + x + 300 = 0

<=> 303 - 99x = 0

<=> 3( 101 - 33x ) = 0

<=> 101 - 33x = 0

<=> x = 101/33

Vậy x = 101/33

3-4x(25-2x)=8x²-x-300

<=>3-100x+8x²=8x²-x-300

<=>3-100x+8x²-8x²+x+300=0

<=>303-99x=0

<=>99x=303

<=>x=101/33

Vậy x=101/33

`b,2(x+1)=5x-7`

`=>2x+2=5x-7`

`=>3x=9`

`=>x=3`

`c,3-4x(25-2x)=8x^2+x-300`

`<=>3-100x+8x^2=8x^2+x-300`

`<=>101x=303`

`<=>x=3`

`d,(10x+3)/12=1+(6+8x)/9`

`<=>(10x+3)/12=(8x+15)/9`

`<=>30x+9=32x+60`

`<=>2x=-51`

`<=>x=-51/2`

a. \(3-4x\left(25-2x\right)-8x^2+x-300=0\)

\(\Leftrightarrow3-100x+8x^2-8x^2+x-300=0\)

\(\Leftrightarrow-297-99x=0\)

\(\Leftrightarrow x=3\)

Vậy \(n_0\) của PT là: x=3

b. \(\Leftrightarrow\frac{\left(2-6x\right)}{5}-2+\frac{3x}{10}=7-\frac{3x+3}{4}\)

\(\Leftrightarrow\frac{\left(4-12x\right)}{5}-\frac{20}{10}+\frac{3x}{10}=\frac{\left(28-3x-3\right)}{4}\)

\(\Leftrightarrow\frac{\left(-16-9x\right)}{10}=\frac{\left(25-3x\right)}{4}\)

\(\Leftrightarrow-64-36x=250-30x\)

\(\Leftrightarrow-6x=314\)

\(\Leftrightarrow x=-\frac{157}{3}\)

Vậy -\(n_0\) của PT là: \(x=\frac{-157}{3}\)

c. \(5x+\frac{2}{6}-8x-\frac{1}{3}=4x+\frac{2}{5}-5\)

\(\Leftrightarrow-3x=4x-\frac{23}{5}\)

\(\Leftrightarrow7x=\frac{23}{5}\)

\(\Leftrightarrow x=\frac{23}{35}\)

Vậy \(n_0\) của PT là: \(x=\frac{23}{35}\)

d. \(3x+\frac{2}{3}-3x+\frac{1}{6}=2x+\frac{5}{3}\)

\(\Leftrightarrow\frac{5}{6}=2x+\frac{5}{3}\)

\(\Leftrightarrow x=-\frac{5}{12}\)

Vậy \(n_0\) của Pt là: \(x=-\frac{5}{12}\)

3-4x(25-2x)= 8x² +x -300

3 - 100x + 8x² = 8x² + x - 300

3 + 300 - 100x - x + 8x² - 8x² = 0

303 -101x = 0

101(3- x)=0

x = 3

a, \(5-\left(x-6\right)=4\left(3-2x\right)\\ \Leftrightarrow5-x+6=12-8x\\ \Leftrightarrow8x-x=12-5-6\\ \Leftrightarrow7x=1\\ \Leftrightarrow x=\frac{1}{7}\)

b, \(3-4x\left(25-2x\right)=8x^2+x-300\\ \Leftrightarrow3-100x+8x^2=8x^2+x-300\\ \Leftrightarrow x+100x=3+300\\ \Leftrightarrow101x=303\\ \Leftrightarrow x=3\)

=>3-100x+8x² = 8x²+x-300

=>3-100x = x-300

=>101x=303

=>x=3

250 do ban

a,  (2x+5)mũ 2=(x+2) mũ 2

=.> (2x+5) mũ 2-(x+2) mũ 2=0

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

=>(3x+7)x(x+3)=0

=>3x+7=0 hoặc x+3=0

3x+7=0=>x=-7/3

x+3=0 =>x=-3

vậy x=-7/3 hoặc x=-3

hok tot