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

\(\Leftrightarrow\left(y-2\right)\left(y-3\right)+\left(y-3\right)=0\)

\(\Leftrightarrow\left(y-3\right)^2=0\)

\(\Leftrightarrow y=3\)

\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

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

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

\(\Leftrightarrow x\in\left\{0;-3;2\right\}\)

Bài làm

Vì ( y - 2 ) . ( y - 3 ) + ( y - 2 ) - 1 = 0

=> ( y - 2 ) = 0 hoặc ( y - 3 ) + ( y - 2 ) - 1 = 0

=> y = 2 hoặc y = 3 

Vậy y = 2 hoặc y = 3

~ Mấy câu còn lại làm tương tự. Làm theo mẫu câu a . b = 0 , => a = 0 hoặc b = 0. ~
# Chúc bạn học tốt # 

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

\(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\)

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

\(\Leftrightarrow x\in\left\{2;-3\right\}\)

\(\left(x-7\right)\left(x+3\right)=\left(x+3\right)\left(2x-9\right)\)

\(\Leftrightarrow\left(x-7\right)\left(x+3\right)-\left(x+3\right)\left(2x-9\right)=0\)

\(\Leftrightarrow\left(x-7-2x+9\right)\left(x+3\right)=0\)

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

\(\Leftrightarrow x\in\left\{2;-3\right\}\)

3^ x -1 = 1/243

3^x =1/243 +1

3^x = 244 / 243

Ta  thấy đây ko phải lũy thừa của 3 => Ko có x thỏa mãn

81^-2x . 27^x =9^5

81^-2 . 81^x . 27^x =9^5

1/9^4 . (81.27)^x =9 ^5

            3^6x = 9^5 : 1/9^4

             3^6x = 9^9

            3^6x = 3^18

    => 6x =18

  x=3

2^x +2^x +3 =144

2.(2^x) =141

2^x+1 = 141

Ta thấy 141 ko phải lũy thừa của 2 => ko có x thỏa mãn

Đề:  Biết  \(8x^3+12x^2y+6xy^2+y^3=27\) . Tính  \(A=x\left(2x+y\right)+xy+\frac{1}{2}y^2\)

                                                     -------------------------

Ta có:

\(8x^3+12x^2y+6xy^2+y^3=27\)

\(\Leftrightarrow\)  \(\left(2x+y\right)^3=27\)

\(\Leftrightarrow\)  \(2x+y=3\)

Do đó:

\(A=3x+xy+\frac{1}{2}y^2\)

\(=3x+\frac{1}{2}y\left(2x+y\right)\)

\(=3x+\frac{3}{2}y\)

\(=\frac{3}{2}\left(2x+y\right)\)

\(A=\frac{9}{2}\)

hic nhìu mà khó nữa *_*

a/ \(x^2+y^2=x^2+y^2+2xy-2xy =\left(x+y\right)^2-2xy\)

b/ mình không chắc nữa

bài 3

a/ \(9x^2-49=0 \Leftrightarrow x^2=\frac{49}{9} \Leftrightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=-\frac{7}{3}\end{cases}}\)

b/ \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x+1\right)\left(x-1\right)-27=0 \Leftrightarrow x^3+27-x\left(x^2-1\right)-27=0\)

\(\Leftrightarrow x^3-x^3+x=0\Leftrightarrow x=0\)

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

\(\Leftrightarrow\left(x+2\right)\left(x-1\right)^2=0\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}}\)

d/ \(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=0\)

\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\)

\(\Leftrightarrow4x+25=0 \Leftrightarrow x=\frac{-25}{4}\)

e/ mình lười qá ko viết đề đâu 

\(\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\)

\(\Leftrightarrow-3x+1=7 \Leftrightarrow x=-2\)

có gì sai bn sửa lại nha 

\(\left(x-3\right)\left(x-12\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=12\end{cases}}\)

\(\Rightarrow x\in\left\{3;12\right\}\)

\(\left(x^2-81\right)\left(x^2+9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2-81=0\\x^2+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x\in\varnothing\end{cases}}\Leftrightarrow x=9\)

\(\Rightarrow x=9\)

\(\left(x-4\right)\left(x+2\right)< 0\)

\(\Rightarrow\hept{\begin{cases}x-4\\x+2\end{cases}}\)trái dấu

\(TH1:\hept{\begin{cases}x-4>0\\x+2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>4\\x< -2\end{cases}}\Leftrightarrow x\in\varnothing\)

\(TH2:\hept{\begin{cases}x-4< 0\\x+2>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 4\\x>-2\end{cases}}\Leftrightarrow x\in\left\{-1;0;1;2;3\right\}\)

Vậy \(x\in\left\{-1;0;1;2;3\right\}\)

a. (x - 2)² = 1

<=> (x - 2)² = 1² = (-1)²

<=> \(\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\begin{cases}x=3\\x=1\end{cases}\)

Vậy x \(\in\){1; 3}.

b. (2x - 1)³ = -8

<=> (2x - 1)³ = (-2)³

<=> 2x - 1 = -2

<=> 2x = -2 + 1

<=> 2x = -1

<=> x = -1/2

Vậy x = -1/2.

c. (x + 1/2)² = 1/16

<=> (x + 1/2)² = (1/4)² = (-1/4)²

<=> \(\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}\)

Vậy x \(\in\){-1/4; -3/4}.

d. (x - 2)³ = -27

<=> (x - 2)³ = (-3)³

<=> x - 2 = -3

<=> x = -3 + 2

<=> x = -1

Vậy x = -1.

a.\(\left(x-2\right)^2\)=1

<=> x-2=1 hoặc x-2=-1

<=> x= 3 hoặc x=1

b.\(\left(2x-1\right)^3\)=-8

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

2x-1=-2

2x=-1

x=-1/2

c.\(\left(x+\frac{1}{2}\right)^2\)=\(\frac{1}{16}\)

\(\left(x+\frac{1}{2}\right)^2\)=\(\left(\frac{1}{4}\right)^2\)hoặc \(\left(x+\frac{1}{2}\right)^2\)=\(\left(-\frac{1}{4}\right)^2\)

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

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

d.\(\left(x-2\right)^3\)=-27

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

x-2=-3

x=-1

g.\(3^{x-1}\)=\(\frac{1}{243}\)

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

\(3^{x-1}\)=\(3^{-5}\)

x-1=-5

x=-4