-5.(x+1/5) -1/2.(x-2/3)=3/2x-5/6

-5x + (-1) -1/2x -1/3=3/2x-5/6

-5x-1/2x-3/2x=1+1/3-5/6

x.(-5-1/2-3/2)= 6/6+2/6+(-5/6)

x.(-10/2+(-1/2)+(-3/2))=3/6

x.6/2=1/2

x=1/2:6/2

x=1/6

Vậy x = 1/6

cách 1:=> (x - 7)^(x+1)= (x-7)^(x+11) 
 

TH1: x-7=0 => x=7 => 0^8=0^18 (TM) 
 

TH2: x-7=1 => x=8 (TM) 
 

TH3: x khác 7 và 8 => x+1=x+11 => vô lý => loại 
 

KL: x = 7 hoặc x=8

( x-7)^( x+1) - ( x-7)^(x+11) = 0 
 

( x-7)^( x+1) - ( x-7)^(x+1)*x^10 = 0 
 

( x-7)^( x+1) (1-x^10) = 0 

tới đây dễ òi

cách 3:\(\Leftrightarrow\left(x-7\right)^{x+1}=\left(x-7\right)^{x+11}\)

\(\Leftrightarrow x-7=0\)hoặc x+1=x+11(vô lí)

\(\Rightarrow x=7\)

a) thay \(x-y=\frac{3}{10}\)vào \(y\left(x-y\right)=\frac{-3}{50}\)ta có\(\frac{3}{10}y=\frac{-3}{50}\)=>\(y=\frac{-3}{50}:\frac{3}{10}=\frac{-1}{5}\)=>\(x-y=\frac{3}{10}\Rightarrow x=\frac{3}{10}+\frac{-1}{5}=\frac{1}{10}\)

hôm sau mik giải tip cho

1.a)\(2.x-\dfrac{5}{4}=\dfrac{20}{15}\)

\(\Leftrightarrow2.x=\dfrac{20}{15}+\dfrac{5}{4}=\dfrac{4}{3}+\dfrac{5}{4}=\dfrac{16+15}{12}=\dfrac{31}{12}\)

\(\Leftrightarrow x=\dfrac{31}{12}:2=\dfrac{31}{12}.\dfrac{1}{2}=\dfrac{31}{24}\)

b)\(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{8}\right)\)

\(\Leftrightarrow\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)

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

\(\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{1}{3}=-\dfrac{5}{6}\)

2.Theo đề bài, ta có: \(\dfrac{a}{2}=\dfrac{b}{3}\) và \(a+b=-15\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{a+b}{2+3}=\dfrac{-15}{5}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=-3\Rightarrow a=-6\\\dfrac{b}{3}=-3\Rightarrow b=-9\end{matrix}\right.\)

3.Ta xét từng trường hợp:

-TH1:\(\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\)\(\Rightarrow x\in\left\{0;1\right\}\)

-TH2:\(\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\)\(\Rightarrow x\in\varnothing\)

Vậy \(x\in\left\{0;1\right\}\)

4.\(B=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^9=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^9=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^{18}=\left(\dfrac{3}{7}\right)^3=\dfrac{27}{343}\)

\(VP=\frac{1}{2\left(a+3\right)}+\frac{1}{2\left(a+5\right)}=\frac{2\left(a+5\right)}{2\left(a+3\right)\left(a+5\right)}+\frac{2\left(a+3\right)}{2\left(a+3\right)\left(a+5\right)}\)

\(=\frac{2\left(a+5\right)}{4\left(a+3\right)\left(a+5\right)}+\frac{2\left(a+3\right)}{4\left(a+3\right)\left(a+5\right)}=\frac{2\left(a+5\right)+2\left(a+3\right)}{4\left(a+3\right)\left(a+5\right)}=\frac{2\left[\left(a+3\right)+\left(a+5\right)\right]}{4\left(a+3\right)\left(a+5\right)}=\frac{\left(a+3\right)+\left(a+5\right)}{2\left(a+3\right)\left(a+5\right)}\)

\(=\frac{\left(a+a\right)+\left(3+5\right)}{2\left(a+3\right)\left(a+5\right)}=\frac{2a+8}{2\left(a+3\right)\left(a+5\right)}=\frac{2\left(a+4\right)}{2\left(a+3\right)\left(a+5\right)}=\frac{a+4}{\left(a+3\right)\left(a+5\right)}\)

\(VT=\frac{x-2}{\left(a+3\right)\left(a-5\right)}\)

\(\Rightarrow\frac{x-2}{\left(a+3\right)\left(a-5\right)}=\frac{a+4}{\left(a+3\right)\left(a+5\right)}\)

\(\Rightarrow\frac{x-2}{a+4}=\frac{\left(a+3\right)\left(a-5\right)}{\left(a+3\right)\left(a+5\right)}\Rightarrow\frac{x-2}{a+4}=\frac{a-5}{a+5}\Rightarrow\left(x-2\right)\left(a+5\right)=\left(a-5\right)\left(a+4\right)\)

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