Ta có: \(5^{2x+1}\cdot5^{x+1}=5^1\)

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

\(\Leftrightarrow2x^2+3x=0\)

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

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

\(\Rightarrow2-4x=6-3x\\ \Rightarrow x=-4\)

\(\dfrac{2x-1}{3}=\dfrac{2-x}{-2}\)

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

\(\Rightarrow-4x+2=6-3x\Rightarrow x=-4\)

Có đúng hông dậy mọi người

`2x+5y=11(1)`

`2x-3y=0(2)`

Lấy (1) trừ (2)

`=>8y=11`

`<=>y=11/8`

`<=>x=(3y)/2=33/16`

a) Ta có: \(\left\{{}\begin{matrix}2x+5y=11\\2x-3y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}8y=11\\2x-3y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{11}{8}\\2x=3y=3\cdot\dfrac{11}{8}=\dfrac{33}{8}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{33}{16}\\y=\dfrac{11}{8}\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{33}{16}\\y=\dfrac{11}{8}\end{matrix}\right.\)

b) Ta có: \(\left\{{}\begin{matrix}4x+3y=6\\2x+y=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x+3y=6\\4x+2y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\2x+y=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-2=4\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=6\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là (x,y)=(3;-2)

`b)4x+3y=6(1)`

`2x+y=4<=>4x+2y=8(2)`

Lấy (1) trừ (2) ta có:

`y=-2`

`<=>x=(4-2y)/2=3`

`-2/5 : x=1/2`

`=> x= -2/5 : 1/2`

`=> x= -2/5 xx 2`

`=>x= -4/5`

__

`7/6 : x = 7/4`

`=>x= 7/6 : 7/4`

`=>x=7/6 xx 4/7`

`=>x= 28/42`

`=>x=2/3`

tìm x hả

\(x=\dfrac{-2}{5}:\dfrac{1}{2}\)

\(x=\dfrac{-4}{5}\)

b. \(x=\dfrac{7}{6}:\dfrac{7}{4}\)

\(x=\dfrac{2}{3}\)

\(|2x^2-3x+4|-|2x-x^2-1|=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}2x^2-3x+4-2x+x^2+1=0\\2x^2-3x+4+2x-x^2-1=0\end{cases}}\)

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

\(\Leftrightarrow\orbr{\begin{cases}3\left(x^2-\frac{5}{3}x+\frac{25}{9}-\frac{25}{9}+\frac{5}{3}\right)=0\\x^2-2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3\left(x-\frac{5}{3}^2\right)-\frac{10}{3}=0\\\left(x-\frac{1}{2}\right)^2+\frac{11}{4}>0\left(Loai\right)\end{cases}}\)

\(\Leftrightarrow\left(x\sqrt{3}-\frac{5\sqrt{3}}{3}\right)^2-\left(\frac{\sqrt{30}}{3}\right)^2=0\)

\(\Leftrightarrow\left(x\sqrt{3}-\frac{5\sqrt{3}}{3}-\frac{\sqrt{30}}{3}\right)\left(x\sqrt{3}-\frac{5\sqrt{3}}{3}+\frac{\sqrt{30}}{3}\right)=0\)

\(\Leftrightarrow\left(x\sqrt{3}-\frac{\sqrt{30}+5\sqrt{3}}{3}\right)\left(x\sqrt{3}+\frac{\sqrt{30}-5\sqrt{3}}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x\sqrt{3}-\frac{\sqrt{30}+5\sqrt{3}}{3}=0\\x\sqrt{3}+\frac{\sqrt{30}-5\sqrt{3}}{3}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{10}}{3}\\x=\frac{5-\sqrt{10}}{3}\end{cases}}\)

Vậy ...

\(\left|2x^2-3x+4\right|-\left|2x-x^2-1\right|=0\)

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

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

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

\(TH1:3x^2-5x+5=0\)

Ta có: \(\Delta=5^2-4.3.5=-35< 0\)(vô nghiệm)

\(TH2:x^2-x+3=0\)

Ta có: \(\Delta=1^2-4.1.3=-11< 0\)(vô nghiệm)

Vậy pt vô nghiệm

Yêu cầu đề?

m mem đề đâu 

g, \(y=\sqrt{-x^2+4x-5}=\sqrt{-\left(x-2\right)^2-1}\)

\(\Rightarrow\) Hàm số này không xác định với mọi x.

h, \(y=\sqrt{x^2+2x+2}=\sqrt{\left(x+1\right)^2+1}>0\forall x\)

\(\Rightarrow\) Hàm số này xác định với mọi x.

i, \(y=\sqrt{\left(x-1\right)\left(x-3\right)}\) xác định khi:

\(\left(x-1\right)\left(x-3\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3\ge0\\x-1\le0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge3\\x\le1\end{matrix}\right.\)