Trả lời:

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

\(\Leftrightarrow\left|2x+1\right|-\left|2x-3\right|=0\)

Vì \(\left|2x+1\right|\ge0\)với \(\forall x\)

    \(\left|2x-3\right|\ge0\)với \(\forall x\)

Do đó: \(\left|2x+1\right|-\left|2x-3\right|\ge0\)với \(\forall x\)

Mà\(\left|2x+1\right|-\left|2x-3\right|=0\)

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

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

\(\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{2}\\x=\frac{3}{2}\end{cases}}\)

\(\Leftrightarrow x\in\varnothing\)

Vậy \(x\in\varnothing\)

Hok tốt!

Vuong Dong Yet

GIÚP MÌNH VỚI MỌI NGƯỜI ƠI

2x+1=2x-3 hoặc 2x+1= - 2x+ 3

suy ra 4x=2

x=1/2

mặc kệ biến chú tâm vào hệ trong ngoặc rồi mũ nó lên

a)1

b)1

\(|5x-3|-x=7\)

\(|5x-3|=7+x\)

\(\orbr{\begin{cases}5x-3=7+x\\5x-3=-7-x\end{cases}}\)

\(\orbr{\begin{cases}5x-x=7+3\\5x+x=-7+3\end{cases}}\)

\(\orbr{\begin{cases}4x=10\\6x=-4\end{cases}}\)

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

Vậy x = 2,5 hoặc x = -2/3

Hi Hi!

Ta có : 3(2x - 1)² \(\ge0\forall x\)

           7(3y + 5)² \(\ge0\forall x\)

Mà : 3(2x - 1)² + 7(3y + 5)² = 0 

Nên : 3(2x - 1)² = 7(3y + 5)² = 0 

\(\Leftrightarrow\hept{\begin{cases}3\left(2x-1\right)^2=0\\7\left(3y+1\right)^2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(2x-1\right)^2=0\\\left(3y+1\right)^2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(2x-1\right)=0\\\left(3y+1\right)=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x=1\\3y=-1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-\frac{1}{3}\end{cases}}\)

1./ \(x+y=3\Rightarrow\left(x+y\right)^3=27\Rightarrow x^3+y^3+3xy\left(x+y\right)=27\Rightarrow x^3+y^3+3\cdot2\cdot3=27.\)

\(\Rightarrow x^3+y^3=9\)

2./ \(\left(x+3\right)\left(x^2-3x+3^2\right)-x^3-2x-4=0\)

\(\Leftrightarrow x^3+27-x^3-2x-4=0\Leftrightarrow2x=23\Leftrightarrow x=\frac{23}{2}\)

1/ \(x+y=3\)

\(\Rightarrow\left(x+y\right)^2=9\)

\(\Rightarrow x^2+2xy+y^2=9\)

\(\Rightarrow x^2+4+y^2=9\)

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

\(\Rightarrow A=x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=3.1=3\)

Bài 1:

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

b) \(-\left(5+4y\right)\left(5-4y\right)=-\left[\left(5+4y\right)\left(5-4y\right)\right]=-\left[5^2-\left(4y\right)^2\right]=-\left(25-16y^2\right)=-25+16y^2\)

a) | \(\frac{1}{2}\)x| = 3 - 2x

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

\(\Rightarrow\orbr{\begin{cases}\frac{5}{2}x=3\\\frac{1}{2}x-2x=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=3:\frac{5}{2}\\-\frac{3}{2}x=-3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{6}{5}\\x=-3:\left(-\frac{3}{2}\right)\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{6}{5}\\x=2\end{cases}}\)

b) |x - 1| = 3x + 2

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

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

\(\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=-\frac{1}{4}\end{cases}}\)

c) | 5x | = x - 12

\(\Rightarrow\orbr{\begin{cases}5x=x-12\\5x=-\left(x-12\right)\end{cases}}\Rightarrow\orbr{\begin{cases}5x-x=-12\\5x=-x+12\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}4x=-12\\5x+x=12\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\6x=12\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

d) |7 - x| = 5x + 1

\(\Rightarrow\orbr{\begin{cases}7-x=5x+1\\7-x=-\left(5x+1\right)\end{cases}}\Rightarrow\orbr{\begin{cases}7-1=5x+x\\7-x=-5x-1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}6=6x\\7+1=-5x+x\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\8=-4x\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

e) |9 + x| = 2x

\(\Rightarrow\orbr{\begin{cases}9+x=2x\\9+x=-2x\end{cases}}\Rightarrow\orbr{\begin{cases}9=2x-x\\9=-2x-x\end{cases}}\Rightarrow\orbr{\begin{cases}9=x\\9=-3x\end{cases}}\Rightarrow\orbr{\begin{cases}x=9\\x=-3\end{cases}}\)

Ủng hộ mk nha !!! ^_^

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

\(x^2+4x+4-3x+3=x^3-3x^2+3x-1-2x+6x^2\)

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

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

\(x^3+2x^2-8=0\)

        Đề bài có sai ko bn