Hỏi đáp Tìm x

\(40+\left(x+6\right)=45\) 

\(x+6=45-40\) 

\(x+6=5\)  

\(x=5-6\) 

\(x=-1\)

 x : 40 + (x+6 ) = 45 

x+6 = 45-40

x+6 = 5 

x= 5-6 

x= -1

40 + ( x + 6 ) = 45

x + 6             = 45 - 40

x + 6             = 5

x                   = 6 - 5

x                   = 1

 \(x:17+5=12\)

\(x:17=12-5\)

\(x:17=7\)

\(x=7\cdot17\)

\(x=119\)

\(x:17=12-5\) 

\(x:17=7\) 

\(x=17\cdot7\) 

\(x=119\)

\(x\div17+5=12\)

\(x\div17=12-5\)

\(x\div17=7\)

\(x=17\times7\)

\(x=119.\)

TH1:x-4=0              TH2:3x-15=0

        x=0+4                      3x=0+15

        x=4                           3x=15

                                           x=15:3

                                           x=5

\(\Rightarrow x-4=0\text{ hoặc }3x-15=0\)

\(\Rightarrow x=4\text{ hoặc }x=5\)

\(\text{Vậy }x=4\text{ hoặc }x=5\)

( x - 4 ) . ( 3x - 15 ) = 0

Thỏa mãn điều kiện x - 4 = 0

x = 4

x - y = xy = x : y 

Xét x - y = xy

=> x = xy + y = y( x + 1 )

Xét x - y = x : y

Thế x = y( x + 1 ) vào VP ta được :

\(x-y=\frac{y\left(x+1\right)}{y}=x+1\)

=> x - y = x + 1 

=> 0 = 1 + y

=> y = -1

Thế y = -1 ta được

=> x + 1 = -x

=> x + x = -1

=> 2x = -1

=> x = -1/2

Vậy x = -1/2 ; y = -1

            Bài làm :

a) x( 2x - 7 ) - 4x + 14 = 0

<=> x( 2x - 7 ) - 2( 2x - 7 ) = 0

<=> ( 2x - 7 )( x - 2 ) = 0

 \(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=2\end{cases}}\)

b) Sửa đề : 5x³ + x² - 4x + 9 = 0

<=>( 5x³ + 5 ) + (x² - 4x +4)=0

<=> 5(x³ + 1) + (x-2)² = 0

<=> 5(x+1)(x² - x +1) + (x+2)² =0

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)

c) 3x³ - 7x² + 6x - 14 = 0

<=> 3x²( x - 7/3 ) + 6( x - 7/3 ) = 0

<=> ( x - 7/3 )( 3x² + 6 ) = 0

 \(\Leftrightarrow\orbr{\begin{cases}x-\frac{7}{3}=0\\3x^2+6=0\end{cases}}\Leftrightarrow x=\frac{7}{3}\)

d) 5x² - 5x = 3( x - 1 )

<=> 5x( x - 1 ) - 3( x - 1 ) = 0

<=> ( x - 1 )( 5x - 3 ) = 0

 \(\Leftrightarrow\orbr{\begin{cases}x-1=0\\5x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{3}{5}\end{cases}}\)

e) 4x² - 25 - ( 4x - 10 ) = 0

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

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

<=> ( 2x - 5 )( 2x + 3 ) = 0

 \(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\2x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)

f) x³ + 27 + ( x + 3 )( x - 9 ) = 0

<=> ( x + 3 )( x² - 3x + 9 ) + ( x + 3 )( x - 9 ) = 0

<=> ( x + 3 )( x² - 3x + 9 + x - 9 ) = 0

<=> ( x + 3 )( x² - 2x ) = 0

<=> x( x + 3 )( x - 2 ) = 0

\(\Leftrightarrow\orbr{\begin{cases}\\\end{cases}}\begin{cases}x=0\\x=-3\\x=2\end{cases}\)

a) x( 2x - 7 ) - 4x + 14 = 0

<=> x( 2x - 7 ) - 2( 2x - 7 ) = 0

<=> ( 2x - 7 )( x - 2 ) = 0

<=> \(\orbr{\begin{cases}2x-7=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=2\end{cases}}\)

b) 5x³ + x² - 4x - 9 = 0 ( đề sai )

c) 3x³ - 7x² + 6x - 14 = 0

<=> 3x²( x - 7/3 ) + 6( x - 7/3 ) = 0

<=> ( x - 7/3 )( 3x² + 6 ) = 0

<=> \(\orbr{\begin{cases}x-\frac{7}{3}=0\\3x^2+6=0\end{cases}}\Leftrightarrow x=\frac{7}{3}\)( do 3x² + 6 ≥ 6 > 0 với mọi x )

d) 5x² - 5x = 3( x - 1 )

<=> 5x( x - 1 ) - 3( x - 1 ) = 0

<=> ( x - 1 )( 5x - 3 ) = 0

<=> \(\orbr{\begin{cases}x-1=0\\5x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{3}{5}\end{cases}}\)

e) 4x² - 25 - ( 4x - 10 ) = 0

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

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

<=> ( 2x - 5 )( 2x + 3 ) = 0

<=> \(\orbr{\begin{cases}2x-5=0\\2x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)

f) x³ + 27 + ( x + 3 )( x - 9 ) = 0

<=> ( x + 3 )( x² - 3x + 9 ) + ( x + 3 )( x - 9 ) = 0

<=> ( x + 3 )( x² - 3x + 9 + x - 9 ) = 0

<=> ( x + 3 )( x² - 2x ) = 0

<=> x( x + 3 )( x - 2 ) = 0

<=> x = 0 hoặc x + 3 = 0 hoặc x - 2 = 0

<=> x = 0 hoặc x = -3 hoặc x = 2

\(x^2-3x+5\left(x-3\right)=0\) 

\(x^2-3x+5x-15=0\) 

\(x^2+2x-15=0\)  

\(x^2-3x+5x-15=0\) 

\(x\left(x-3\right)+5\left(x-3\right)=0\) 

\(\left(x+5\right)\left(x-3\right)=0\) 

\(\orbr{\begin{cases}x+5=0\\x-3=0\end{cases}}\) 

\(\orbr{\begin{cases}x=-5\\x=3\end{cases}}\)

\(x^2-3x+5\left(x-3\right)=0\)

\(\Leftrightarrow x\left(x-3\right)+5\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)

Vậy \(x=3\)hoặc \(x=-5\)

x² - 3x + 5( x - 3 ) = 0

<=> x( x - 3 ) + 5( x - 3 ) = 0

<=> ( x - 3 )( x + 5 ) = 0

<=> \(\orbr{\begin{cases}x-3=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)

(x - 5)(2x - 2) + (x - 5) = 0  

<=> 2x² - 2x - 10x + 10 + x - 5 = 0

<=> 2x² - 11x + 5 = 0 

<=> 2x² - x - 10x + 5 = 0

<=> x (2x - 1) - 5 (2x - 1) = 0

<=> (x - 5)(2x - 1) = 0

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\2x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{2}\end{cases}}\)

Vậy x \(\in\left\{5;\frac{1}{2}\right\}\)

Học tốt nhá :))

              Bài làm :

Ta có :

\(...\)

\(\Leftrightarrow\left(x-5\right)\left(2x-2+1\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\2x-1\end{cases}=0}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=0,5\end{cases}}\)

Vậy x=5 hoặc x= 0,5

Bài làm :

\(\left(x-5\right)\left(2x-2\right)+\left(x-5\right)=0\)

\(\left(x-5\right)\left(2x-2+1\right)=0\)

\(\left(x-5\right)\left(2x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\2x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\2x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{2}\end{cases}}\)

Vậy x = 5 hoặc \(x=\frac{1}{2}\) .

Học tốt

a) \(\left|2x-1\right|+\frac{1}{3}=0\)

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

=> vô lý

=> PT vô nghiệm

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

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

Vì \(\hept{\begin{cases}\left|x+2\right|\ge0\\-\left|x-3\right|\le0\end{cases}\left(\forall x\right)}\) nên dấu "=" xảy ra khi: 

\(\left|x+2\right|=-\left|x-3\right|=0\Rightarrow\hept{\begin{cases}x=-2\\x=3\end{cases}}\) (vô lý)

=> PT vô nghiệm

(99999-x)+10204=2340249

99999-x=2340249-10204=2330045

x=99999-2330045=-2230046

Không có tổng mà tìm x thì suy ra đpcm.

( x - 1 )( x² + x + 1 ) - ( x + 2 )( x - 2 )x = 39

<=> x³ - 1 - ( x² - 4 )x = 39

<=> x³ - 1 - x² + 4x = 39

<=> 4x - 1 = 39

<=> 4x = 40

<=> x = 40

Bài làm :

\(\left(x-1\right)\left(x^2+x+1\right)-\left(x+2\right)\left(x-2\right)x=39\)

\(\Leftrightarrow x^3+x^2+x-x^2-x-1-\left(x^2-4\right)x=39\)

\(\Leftrightarrow x^3-1-x^3+4x=39\)

\(\Leftrightarrow\left(x^3-x^3\right)+4x=39+1\)

\(\Leftrightarrow4x=40\)

\(\Leftrightarrow x=10\)

Vậy x = 10 .

Học tốt nhé

         Bài làm :

Ta có :

(x - 1)(x² + x + 1) - (x + 2)(x - 2)x = 39

<=> x³ - 1 - (x² - 4).x = 39

<=> x³ - 1 - x³ + 4x = 39

<=> 4x = 40

<=> x = 10

Vậy x = 10

           Bài làm :

Ta có :

(3x-6).(4x-8)=0

\(\Rightarrow\orbr{\begin{cases}3x-6=0\Rightarrow x=2\\4x-8=0\Rightarrow x=2\end{cases}}\)

Vậy x=2

Bài làm :

\(\left(3x-6\right).\left(4x-8\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-6=0\\4x-8=0\end{cases}}\Rightarrow\orbr{\begin{cases}3x=6\\4x=8\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=2\end{cases}}\)

Vậy x = 2 .

Học tốt

\(\left(3x-6\right)\cdot\left(4x-8\right)=0\) 

\(\orbr{\begin{cases}3x-6=0\\4x-8=0\end{cases}}\) 

\(\orbr{\begin{cases}3x=6\\4x=8\end{cases}}\) 

\(\orbr{\begin{cases}x=2\\x=2\end{cases}}\) 

\(x=2\) 

BAI KHO

            Bài làm :

Ta có :

x(2x + 3) - (x + 2)(x + 1) = 6

<=> 2x² + 3x - (x² + 3x + 2) = 6

<=> 2x² + 3x - x² - 3x - 2 = 6

<=> x² = 8

\(\Leftrightarrow x=\pm\sqrt{8}\)

\(\text{Vậy}:x=\pm\sqrt{8}\)

x( 2x + 3 ) - ( x + 2 )( x + 1 ) = 6

⇔ 2x² + 3x - ( x² + 3x + 2 ) = 6

⇔ 2x² + 3x - x² - 3x - 2 = 6

⇔ x² - 2 = 6

⇔ x² = 8

⇔ x = ±√8

             Bài làm :

a) x⁴ - 16x² = 0

<=> ( x² )² - ( 4x )² = 0

<=> ( x² - 4x )( x² + 4x ) = 0

<=> [ x( x - 4 ) ][ x( x + 4 ) ] = 0

<=> x( x - 4 )x( x + 4 ) = 0

<=> x²( x - 4 )( x + 4 ) = 0

 Vậy x=0 hoặc x=±4

b) 9x² + 6x + 1 = 0

<=> ( 3x )² + 2.3x.1 + 1² = 0

<=> ( 3x + 1 )² = 0

<=> 3x + 1 = 0

<=> 3x = -1

<=> x = -1/3

c) x² - 6x = 16

<=> x² - 6x - 16 = 0

<=> x² + 2x - 8x - 16 = 0

<=> x( x + 2 ) - 8( x + 2 ) = 0

<=> ( x + 2 )( x - 8 ) = 0

 \(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}\)

d) 9x² + 6x = 80

<=> 9x² + 6x - 80 = 0

<=> 9x² + 30x - 24x - 80 = 0

<=> 9x( x + 10/3 ) - 24( x + 10/3 ) = 0

<=> ( x + 10/3 )( 9x - 24 ) = 0

 \(\Leftrightarrow\orbr{\begin{cases}x+\frac{10}{3}=0\\9x-24=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{10}{3}\\x=\frac{8}{3}\end{cases}}\)

e) 25( 2x - 1 )² - 9( x + 1 )² = 0

<=> 5²( 2x - 1 )² - 3²( x + 1 )² = 0 

<=> [ 5( 2x - 1 ) ]² - [ 3( x + 1 ) ]² = 0

<=> ( 10x - 5 )² - ( 3x + 3 )² = 0

<=> [ ( 10x - 5 ) - ( 3x + 3 ) ][ ( 10x - 5 ) + ( 3x + 3 ) ] = 0

<=> ( 10x - 5 - 3x - 3 )( 10x - 5 + 3x + 3 ) = 0

<=> ( 7x - 8 )( 13x - 2 ) = 0

 \(\Leftrightarrow\orbr{\begin{cases}7x-8=0\\13x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{7}\\x=\frac{2}{13}\end{cases}}\)

a) x⁴ - 16x² = 0

<=> ( x² )² - ( 4x )² = 0

<=> ( x² - 4x )( x² + 4x ) = 0

<=> [ x( x - 4 ) ][ x( x + 4 ) ] = 0

<=> x( x - 4 )x( x + 4 ) = 0

<=> x²( x - 4 )( x + 4 ) = 0

<=> \(\hept{\begin{cases}x^2=0\\x-4=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)( thay bằng dấu hoặc hộ mình nhé )

b) 9x² + 6x + 1 = 0

<=> ( 3x )² + 2.3x.1 + 1² = 0

<=> ( 3x + 1 )² = 0

<=> 3x + 1 = 0

<=> 3x = -1

<=> x = -1/3

c) x² - 6x = 16

<=> x² - 6x - 16 = 0

<=> x² + 2x - 8x - 16 = 0

<=> x( x + 2 ) - 8( x + 2 ) = 0

<=> ( x + 2 )( x - 8 ) = 0

<=> \(\orbr{\begin{cases}x+2=0\\x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}\)

d) 9x² + 6x = 80

<=> 9x² + 6x - 80 = 0

<=> 9x² + 30x - 24x - 80 = 0

<=> 9x( x + 10/3 ) - 24( x + 10/3 ) = 0

<=> ( x + 10/3 )( 9x - 24 ) = 0

<=> \(\orbr{\begin{cases}x+\frac{10}{3}=0\\9x-24=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{10}{3}\\x=\frac{8}{3}\end{cases}}\)

e) Áp dụng công thức aⁿ.bⁿ = ( ab )ⁿ ta có :

25( 2x - 1 )² - 9( x + 1 )² = 0

<=> 5²( 2x - 1 )² - 3²( x + 1 )² = 0 

<=> [ 5( 2x - 1 ) ]² - [ 3( x + 1 ) ]² = 0

<=> ( 10x - 5 )² - ( 3x + 3 )² = 0

<=> [ ( 10x - 5 ) - ( 3x + 3 ) ][ ( 10x - 5 ) + ( 3x + 3 ) ] = 0

<=> ( 10x - 5 - 3x - 3 )( 10x - 5 + 3x + 3 ) = 0

<=> ( 7x - 8 )( 13x - 2 ) = 0

<=> \(\orbr{\begin{cases}7x-8=0\\13x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{7}\\x=\frac{2}{13}\end{cases}}\)

a) Ta có : x⁴ - 16x² = 0

=> x⁴ - 8x² - 8x² + 64 = 64

=> x²(x² - 8) - 8(x² - 8) = 64

=> (x² - 8)² = 64

=> \(\orbr{\begin{cases}x^2-8=8\\x^2-8=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=16\\x^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\pm4\\x=0\end{cases}}\Rightarrow x\in\left\{4;-4;0\right\}\)

b) Ta có 9x² + 6x + 1 = 0

=> 9x² + 3x + 3x + 1 = 0

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

=> (3x + 1)² = 0

=> 3x + 1 = 0

=> x = -1/3

c) Ta có x² - 6x = 16

=> x² - 6x + 9 = 25

=> (x - 3)² = 25

=> \(\orbr{\begin{cases}x-3=5\\x-3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=-2\end{cases}}\Rightarrow x\in\left\{8;-2\right\}\)

d) 9x² + 6x = 80

=> 9x² + 6x + 1 = 81

=> (3x + 1)² = 81

=> \(\orbr{\begin{cases}3x+1=9\\3x+1=-9\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{8}{3}\\x=-\frac{10}{3}\end{cases}\Rightarrow x\in}\left\{\frac{8}{3};\frac{-10}{3}\right\}\)

e) 25(2x - 1)² - 9(x + 1)² = 0

=> [5(2x - 1)]² - [3(x + 1)]² = 0

=> (10x - 5)² - (3x + 3)² = 0

=> (10x - 5 - 3x - 3)(10x - 5 + 3x + 3) = 0

=> (7x - 8)(13x - 2) = 0

=> \(\orbr{\begin{cases}7x=8\\13x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{8}{7}\\x=\frac{2}{13}\end{cases}}\)

            Bài làm :

Ta có :

...

<=> 3,2 X = 2,4-1,7

<=. X =0,7/3,2=7/32

Vậy X=7/32

\(x\times1,2+x+1,7+x=2,4\)

=> \(x\times1,2+2\times x+1,7=2,4\)

=> \(x\times\left(1,2+2\right)=2,4-1,7\)

=> \(x\times3,2=0,7\)

=> \(x=0,7:3,2=\frac{7}{32}\)

\(2x+\frac{3}{7}=\frac{17}{7}\Leftrightarrow2x=\frac{14}{7}\Leftrightarrow x=1\)

\(\left|2x+1\right|=3\)

TH1 : \(2x+1=3\Leftrightarrow x=1\)

TH2 : \(2x+1=-3\Leftrightarrow x=-2\)

\(2x+\frac{3}{7}=\frac{17}{7}=>2x=\frac{17}{7}-\frac{3}{7}=>2x=2=>x=2:2=>x=1\)

vậy \(x=1 \)

\(\left|2x+1\right|=3=>\orbr{\begin{cases}\left|2x+1\right|=3\\\left|2x+1\right|=-3\end{cases}}=>\orbr{\begin{cases}2x=3+1\\2x=-3+1\end{cases}}=>\orbr{\begin{cases}2x=4\\2x=-2\end{cases}}=>\orbr{\begin{cases}x=4:2\\x=-2:2\end{cases}}=>\orbr{\begin{cases}x=2\\x=1\end{cases}}\)

vậy \(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

a) 

\(2x=\frac{17}{7}-\frac{3}{7}\) 

\(2x=\frac{14}{7}\) 

\(2x=2\) 

\(x=1\)

b) 

\(\orbr{\begin{cases}2x+1=3\\2x+1=-3\end{cases}}\) 

\(\orbr{\begin{cases}2x=2\\2x=-4\end{cases}}\) 

\(\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

\(\left(\frac{2}{3}.x+\frac{5}{6}\right).\left(\frac{3}{4}.x-\frac{27}{8}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}.x+\frac{5}{6}=0\\\frac{3}{4}.x-\frac{27}{8}=0\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{2}{3}.x=\frac{-5}{6}\\\frac{3}{4}.x=\frac{27}{8}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-5}{4}\\x=\frac{9}{2}\end{cases}}\)

Vậy \(x=\frac{-5}{4}\) hoặc \(x=\frac{9}{2}\).

Học tốt

                Bài làm :

\(\left(\frac{2}{3}x+\frac{5}{6}\right).\left(\frac{3}{4}x-\frac{27}{8}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{3}x+\frac{5}{6}=0\\\frac{3}{4}x-\frac{27}{8}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{3}x=\frac{-5}{6}\\\frac{3}{4}x=\frac{27}{8}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-5}{4}\\x=\frac{9}{2}\end{cases}}\)

Vậy x=-5/4 hoặc x=9/2

\(\left(\frac{2}{3}.x+\frac{5}{6}\right).\left(\frac{3}{4}.x-\frac{27}{8}\right)=0=>\orbr{\begin{cases}\left(\frac{3}{4}.x-\frac{27}{8}\right)=0\\\left(\frac{2}{3}.x+\frac{5}{6}\right)=0\end{cases}}\)

\(\orbr{\begin{cases}\frac{2}{3}.x=0-\frac{5}{6}\\\frac{3}{4}.x=0+\frac{27}{8}\end{cases}}=>\orbr{\begin{cases}\frac{2}{3}.x=-\frac{5}{6}\\\frac{3}{4}.x=\frac{27}{8}\end{cases}}=>\orbr{\begin{cases}x=-\frac{5}{6}:\frac{2}{3}\\x=\frac{27}{8}:\frac{3}{4}\end{cases}}=>\orbr{\begin{cases}x=-\frac{5}{4}\\x=\frac{9}{2}\end{cases}}\)

vậy \(\orbr{\begin{cases}x=-\frac{5}{4}\\x=\frac{9}{2}\end{cases}}\)