3 nhân 7^(2x-4)-2=7^(2x-4) làm nhanh lên giùm nha xin đó
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Bài 1:
\(A=\frac{5}{3.6}+\frac{5}{6.9}+....+\frac{5}{96.99}\)
\(\Rightarrow\frac{3}{5}A=\frac{3}{3.6}+\frac{3}{6.9}+....+\frac{3}{96.99}\)
\(\Rightarrow\frac{3}{5}A=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{96}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(\Rightarrow A=\frac{32}{99}\div\frac{3}{5}=\frac{160}{297}\)
Bái 2:
\(B=\frac{2}{3.7}+\frac{2}{7.11}+...+\frac{2}{99.103}\)
\(\Rightarrow2B=\frac{4}{3.7}+\frac{4}{7.11}+....+\frac{4}{99.103}\)
\(\Rightarrow2B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{99}-\frac{1}{103}\)
\(=\frac{1}{3}-\frac{1}{103}=\frac{100}{309}\)
\(\Rightarrow B=\frac{100}{309}\div2=\frac{50}{309}\)
Bài 1:
Ta có:
\(\frac{5}{n.\left(n+3\right)}=\frac{5}{3}.\frac{3}{n.\left(n+3\right)}=\frac{5}{3}.\frac{\left(n+3\right)-n}{n.\left(n+3\right)}=\frac{5}{3}.\left[\frac{n+3}{n.\left(n+3\right)}-\frac{n}{n\left(n+3\right)}\right]\)\(=\frac{5}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)\)
\(\frac{5}{3.6}+\frac{5}{6.9}+\frac{5}{9.12}+...+\frac{5}{96.99}=\frac{5}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{96}-\frac{1}{99}\right)\)
a) 70-5(x-3) = 45
5(x-3) = 70 - 45
5(x-3) = 25
x-3 = 25 : 5
x-3 = 5
x = 5 + 3
x = 8
b) 10+2x = 45 : 43
10+2x = 42
10+2x = 16
2x = 16 - 10
2x = 6
x = 6 : 2
x = 3
Hot boy!~~
a) 70-5.(x-3)=45
=> 5.(x-3)=70-45
=>5.(x-3)=25
=>x-3=25:5
=>x-3=5
=>x=8
b) \(10+2x=4^5:4^3\)
\(\Rightarrow10+2x=16\)
\(\Rightarrow2x=16-10\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
a) \(\left(x-3\right)^2-4=0\)
\(\left(x-7\right)\left(x+1\right)=0\)
\(\orbr{\begin{cases}x=7\\x=-1\end{cases}}\)
b) \(x^2-2x=24\)
\(x^2-2x-24=0\)
\(\left(x-6\right)\left(x+4\right)=0\)
\(\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)
c) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(4x^2+4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(5x^2+10x+10-5x^2+245=0\)
\(10x+255=0\)
\(x=-25.5\)
A) \(\left(x-3\right)^2-4=0\)
\(\left(x-3\right)^2=4\Rightarrow\left(x-3\right)^2=\left(-2\right)^2;2^2\)
th1\(\left(x-3\right)^2=2^2\)
\(\Rightarrow x-3=2\)
\(\Rightarrow x=2+3\)
\(\Rightarrow x=5\)
th2: \(\left(x-3\right)^2=\left(-2\right)^2\)
\(\Rightarrow x-3=-2\)
\(\Rightarrow x=-2+3\)
\(\Rightarrow x=1\)
\(\Leftrightarrow x\in\left\{1;5\right\}\)
a, ( x + 1 ) . ( y + 2 ) = 4
Vì x,y là số tự nhiên nên:
TH1: \(\hept{\begin{cases}x+1=1\\y+2=4\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0\\y=2\end{cases}}\)
TH2: \(\hept{\begin{cases}x+1=2\\y+2=2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=0\end{cases}}\)
b , ( 2x - 1 ) . ( y + 1 ) = 7
Vì x,y là số tự nhiên nên:
TH1: \(\hept{\begin{cases}2x-1=1\\y+1=7\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=6\end{cases}}\)
TH2: \(\hept{\begin{cases}2x-1=7\\y+1=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\\y=0\end{cases}}}\)
c , x + 6 = y . ( x - 1 )
\(\Leftrightarrow x-xy+y+6=0\)
\(\Leftrightarrow-x\left(y-1\right)+\left(y-1\right)=-7\)
\(\Leftrightarrow\left(y-1\right)\left(x-1\right)=7\)
Vì x,y là số tự nhiên nên:
TH1: \(\hept{\begin{cases}y-1=1\\x-1=7\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=8\end{cases}}}\)
TH2: \(\hept{\begin{cases}y-1=7\\x-1=1\end{cases}\Leftrightarrow}\hept{\begin{cases}y=8\\x=2\end{cases}}\)
d, 2xy + 6x + y = 1
\(\Leftrightarrow2x\left(y+3\right)+\left(y+3\right)=4\)
\(\Leftrightarrow\left(2x+1\right)\left(y+3\right)=4\)
Vì x,y là số tự nhiên nên::
\(\hept{\begin{cases}2x+1=1\\y+3=4\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\y=1\end{cases}}}\)
3 . 7(2x-4 ) - 2 = 7(2x-4)
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