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a) | 9 + 7x | = 3 - 5x
\(\Rightarrow\orbr{\begin{cases}9+7x=3-5x\\9+7x=-\left(3-5x\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}7x+5x=3-9\\9+7x=-3+5x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}12x=-6\\7x-5x=-3-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{6}\\2x=-12\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{6}\\x=-6\end{cases}}\)
Bài 2:
1: =>5x+1=6/7 hoặc 5x+1=-6/7
=>5x=-1/7 hoặc 5x=-13/7
=>x=-1/35 hoặc x=-13/35
2: =>x-1=4
=>x=5
3: =>3x-1=3
=>3x=4
=>x=4/3
4: \(\Leftrightarrow\dfrac{5}{x+3}=\dfrac{-5}{6}+\dfrac{1}{2}=\dfrac{-5+3}{6}=\dfrac{-2}{6}=\dfrac{-1}{3}\)
=>x+3=-15
=>x=-18
7: \(\Leftrightarrow2^{2x+1}+2^{2x+6}=264\)
=>2^2x+1*(1+2^5)=264
=>2^2x+1=8
=>2x+1=3
=>x=1
9: =>x^4=8x
=>x^4-8x=0
=>x=2
Bài 7 :
\(\frac{1}{4}-\left(2x-1\right)^2=0\)
\(\left(2x-1\right)^2=\frac{1}{4}-0\)
\(\left(2x-1\right)^2=\frac{1}{4}\)
\(\left(2x-1\right)^2=\left(\frac{1}{2}\right)^2\)
TH1:\(\Rightarrow2x-1=\frac{1}{2}\)
\(2x=\frac{1}{2}+1\)
\(2x=\frac{3}{2}\)
\(x=\frac{3}{4}\)
TH2:\(\Rightarrow2x-1=-\frac{1}{2}\)
\(2x=-\frac{1}{2}+1\)
\(2x=\frac{1}{2}\)
\(x=\frac{1}{4}\)
Vậy x \(\in\left\{\frac{1}{4};\frac{3}{4}\right\}\)
Bài 6 :
\(3^{x+1}=81\)
\(3^{x+1}=3^4\)
\(x+1=4\)
\(\Rightarrow x=3\)
Vậy x = 3
e)
=> (x-2) . (x+7) = ( x-1 ) . ( x +4)
=> x2 +7x - 2x -14 = x2 - x + 4x - 4
x2 + 5x - 14 = x2 + 3x - 4
=> 5x - 14 = 3x - 4
=> 5x - 3x = 14-4
=> 2x = 10 => x = 10 : 2 => x = 5
c)
=>( x-1) . 7 = ( x + 5 ) . 6
=> 7x - 7 = 6x + 30
=> 7x - 6x= 30 + 7
=> x = 37
a,x=\(\frac{5}{2}\)
b,x=\(\frac{13}{176}\)
c,x=37
d, x=\(\frac{12}{5}\)
e, x=5
a, =\(6\cdot\left(-2\right)^3-\left(-2\right)^{10}+4\cdot\left(-2\right)^3+\left(-2\right)^{10}-8\cdot\left(-2\right)^3+\left(-2\right)\)
= \(\left(-48\right)-1024+\left(-32\right)+1024-\left(-64\right)+\left(-2\right)\)
= \(\left(-18\right)\)
b, = \(4\cdot1^6\cdot\left(-1\right)^3-3\cdot1^6\cdot\left(-1\right)^3+2\cdot1^2\cdot\left(-1\right)^2-1^6\cdot\left(-1\right)^3-1^2\cdot\left(-1\right)^2+\left(-1\right)\)
= \(\left(-4\right)-\left(-3\right)+2-\left(-1\right)-1+\left(-1\right)\)
= 0
Bài 7:
x/1=z/2 nên x/6=z/12
=>x/6=y/9=z/12
=>x/2=y/3=z/4
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{27}{9}=3\)
=>x=6; y=9; z=12
\(\Leftrightarrow6^x\cdot\dfrac{1}{6}+6^x\cdot36=6^7\cdot217\)
\(\Leftrightarrow6^x=1679616\)
hay x=8