a) 4(x+2) - 7(2x - 1) + 9(3x - 4)=30

⇔4x+8 - 14x + 7 + 27x - 36 = 30

⇔ 17x = 51

⇔ x = 3

b) 2(5x - 8) - 3(4x - 5) = 4(3x - 4) + 11

⇔ 10x - 16 - 12x + 15 = 12x - 16 + 11

⇔ -14x = -4

⇔ x= \(\frac{2}{7}\)

c) 5x(1 - 2x) - 3x(x + 18) = 0

⇔ 5x - 10x\(^2\) - 3x\(^2\) -54x =0

⇔ -13x\(^2\) -49 x = 0

⇔ -x ( 13x + 49 ) =0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\13x+49=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-49}{13}\end{matrix}\right.\)

d) 5x - 3{4x - 2[4x - 3(5x - 2)]} = 182

⇔ 5x - 3[ 4x - 2( 4x - 15x + 6 ) ]= 182

⇔5x - 3 ( 4x - 8x + 30x - 12 ) = 182

⇔ 5x - 3 ( 26x - 12 ) = 182

⇔ 5x - 78x + 36 = 182

⇔ - 73x = 146

⇔ x = -2

\(|-2x+1,5|=\dfrac{1}{4}\Rightarrow-2x+1,5=\pm\dfrac{1}{4}\)

\(-2x+1,5=\dfrac{1}{4}\Rightarrow-2x=1,5-0,25\Rightarrow-2x=1,25\Rightarrow x=1,25:\left(-2\right)\Rightarrow x=...\)

\(-2x+1,5=-\dfrac{1}{4}\Rightarrow-2x=-0,25-1,5\Rightarrow-2x=1,75\Rightarrow x=1,75:\left(-2\right)\Rightarrow x=...\)

\(\dfrac{3}{2}-|1.\dfrac{1}{4}+3x|=\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{3}{2}-\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{5}{4}\)

\(\Rightarrow1.\dfrac{1}{4}+3x=\pm\dfrac{5}{4}\)

\(1.\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow3x=\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=1\Rightarrow x=3\)

\(1.\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow3x=-\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=-\dfrac{3}{2}x=...\)

a: ta có: \(\left|-2x+\dfrac{3}{2}\right|=\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x+\dfrac{3}{2}=\dfrac{1}{4}\\-2x+\dfrac{3}{2}=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=-\dfrac{5}{4}\\-2x=-\dfrac{7}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{7}{8}\end{matrix}\right.\)

b: Ta có: \(\dfrac{3}{2}-\left|\dfrac{5}{4}+3x\right|=\dfrac{1}{4}\)

\(\Leftrightarrow\left|3x+\dfrac{5}{4}\right|=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{5}{4}=\dfrac{5}{4}\\3x+\dfrac{5}{4}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=0\\3x=-\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{6}\end{matrix}\right.\)

Mấy bài này căng vậy?

a)4(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)

<=>72 - 20x - 36x +84 = 30x - 240 - 6x 84

<=> -80x = -480

<=> x = 6

b) 5(3x+5)-4(2x-3) =5x+3(2x+12)+1

<=> 15x + 25  - 8x + 12 = 5x + 6x + 36 + 1

<=> 15x + 25 - 8x + 12 - 5x - 6x - 36 - 1 = 0

<=> -4x = 0

<=> x = 0

c) 2(5x-8)-3(4x-5)=4(3x-4)+11

= 10x - 16 - 12x + 15 = 12x - 16 + 11

= -14x = -4

= x =\(\frac{2}{7}\)

d) 5x-3{4x-2[4x-3(5x-2)]}=182

= 5x - 3 . [4x - 2(4x - 15x + 6)]

= 5x - 3 . (4x - 8x + 30x - 12)

= 5x - 12x + 24x - 90x + 36

= -73x + 36 = 182

=> -73x = 182 - 36 = 146

=> x = 146 : (-73) = -2

~Hok tốt~

a)4(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)

<=>72 - 20x - 36x +84 = 30x - 240 - 6x 84

<=> -80x = -480

<=> x = 6

b) 5(3x+5)-4(2x-3) =5x+3(2x+12)+1

<=> 15x + 25  - 8x + 12 = 5x + 6x + 36 + 1

<=> 15x + 25 - 8x + 12 - 5x - 6x - 36 - 1 = 0

<=> -4x = 0

<=> x = 0

c) 2(5x-8)-3(4x-5)=4(3x-4)+11

= 10x - 16 - 12x + 15 = 12x - 16 + 11

= -14x = -4

= x = 2/7

d) 5x-3{4x-2[4x-3(5x-2)]}=182

= 5x - 3 . [4x - 2(4x - 15x + 6)]

= 5x - 3 . (4x - 8x + 30x - 12)

= 5x - 12x + 24x - 90x + 36

= -73x + 36 = 182

=> -73x = 182 - 36 = 146

=> x = 146 : (-73) = -2

a) \(\frac{3}{4}x-\frac{1}{4}=2\left(x-3\right)+\frac{1}{4}x\)

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

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

\(x\left(\frac{3}{4}-2-\frac{1}{4}\right)=-\frac{23}{4}\)

\(-\frac{3}{2}x=-\frac{23}{4}\)

\(x=-\frac{23}{4}\div\left(-\frac{3}{2}\right)\)

\(x=\frac{23}{6}\)

b) \(30\%-x+\frac{5}{6}=\frac{1}{3}\)

\(\frac{3}{10}-x+\frac{5}{6}=\frac{1}{3}\)

\(\frac{3}{10}-x=\frac{1}{3}-\frac{5}{6}\)

\(\frac{3}{10}-x=-\frac{1}{2}\)

\(x=\frac{3}{10}-\left(-\frac{1}{2}\right)\)

\(x=\frac{4}{5}\)

c) \(\left(2x-1\right).\left(3x+12\right)=0\)

\(\hept{\begin{cases}2x-1=0\\3x+12=0\end{cases}\Rightarrow\hept{\begin{cases}2x=1\\3x=-12\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{2}\\x=-4\end{cases}}}\)

Vậy \(x\in\left\{\frac{1}{2};-4\right\}\)

a) Ta có: \(x^2+4x+3\)

\(=x^2+x+3x+3\)

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

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

b) Ta có: \(16x-5x^2-3\)

\(=-5x^2+16x-3\)

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

\(=-5x\left(x-3\right)+\left(x-3\right)\)

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

c) Ta có: \(2x^2+7x+5\)

\(=2x^2+2x+5x+5\)

\(=2x\left(x+1\right)+5\left(x+1\right)\)

\(=\left(x+1\right)\left(2x+5\right)\)

d) Ta có: \(2x^2+3x-5\)

\(=2x^2+5x-2x-5\)

\(=x\left(2x+5\right)-\left(2x+5\right)\)

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

e) Ta có: \(x^3-3x^2+1-3x\)

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

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

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

f) Ta có: \(x^2-4x-5\)

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

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

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

\(=\left(x-5\right)\left(x+1\right)\)

g) Ta có: \(\left(a^2+1\right)^2-4a^2\)

\(=\left(a^2+1\right)^2-\left(2a\right)^2\)

\(=\left(a^2+1-2a\right)\left(a^2+1+2a\right)\)

\(=\left(a-1\right)^2\cdot\left(a+1\right)^2\)

h) Ta có: \(x^3-3x^2-4x+12\)

\(=x^2\left(x-3\right)-4\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-4\right)\)

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

i) Ta có: \(x^4+x^3+x+1\)

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

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

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

k) Ta có: \(x^4-x^3-x^2+1\)

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

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

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

l) Ta có: \(\left(2x+1\right)^2-\left(x-1\right)^2\)

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

\(=3x\left(x+2\right)\)

m) Ta có: \(x^4+4x^2-5\)

\(=x^4-x^2+5x^2-5\)

\(=x^2\left(x^2-1\right)+5\left(x^2-1\right)\)

\(=\left(x^2-1\right)\left(x^2+5\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)

thứ nhất bn đăng sai môn 

thứ hai bn giải r đăng lmj :???

Thứ nhất đang sai môn 

Thứ hai không biết giải fndf]-0jhdfuhiofghjfgoihjfgopihjfgihjohjgo;hjghghgdjhldhjdfighjs;dligjlkdfgjdhfghfgh41fg6j541fg3j5h4gf6j54dgh65gf4654j

5gj5fg

35j4gh

6jfd4

5j4fj

😱😱😱😱😱😱😱😱😱😱😱