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\(4.3^x+3^{x+1}=63\)
\(\Rightarrow4.3^x+3.3^x=63\)
\(\Rightarrow7.3^x=63\Rightarrow3^x=9=3^2\Rightarrow x=2\)
\(9.\left(\dfrac{2}{3}\right)^{x+2}-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)
\(\Rightarrow9.\left(\dfrac{2}{3}\right)^2\left(\dfrac{2}{3}\right)^x-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)
\(\Rightarrow9.\dfrac{4}{9}^{ }.\left(\dfrac{2}{3}\right)^x-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x.\left(4-1\right)=\dfrac{4}{3}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x.\dfrac{1}{3}=\dfrac{4}{3}\Rightarrow\left(\dfrac{2}{3}\right)^x=4\)
mà \(0< \left(\dfrac{2}{3}\right)^x< 1;4>0;x>0\)
\(\Rightarrow x\in\varnothing\)
Từ tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{3x+2}{3}=\dfrac{2y-6}{9}=\dfrac{\left(3x+2\right)+\left(2y-6\right)}{3+9}=\dfrac{3x+2y-4}{12}=\dfrac{3x+2y-4}{6x}\)
Suy ra 6x = 12 <=> x = 12 : 6 = 2
Khi đó \(\dfrac{3x+2}{3}=\dfrac{3\cdot2+2}{3}=\dfrac{8}{3}\)
Suy ra \(\dfrac{2y-6}{9}=\dfrac{8}{3}\Leftrightarrow2y-6=\dfrac{8\cdot9}{3}=24\)
\(\Leftrightarrow2y=24+6=30\Leftrightarrow y=30:2=15\)
Vậy x = 2; y = 15
\(x^2-\dfrac{1}{5}.\dfrac{5}{4}=\dfrac{3}{4}\\ \Rightarrow x^2-\dfrac{1}{4}=\dfrac{3}{4}\\ \Rightarrow x^2=1\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
\(\left(\dfrac{1}{2}-x\right)^2=\dfrac{1}{9}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}-x=-\dfrac{1}{3}\\\dfrac{1}{2}-x=\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}\\x=\dfrac{1}{6}\end{matrix}\right.\)
\(x^2-5=5\\ \Rightarrow x^2=10\\ \Rightarrow x=\pm\sqrt{10}\)
\(3x^2-1=14\\ \Rightarrow3x^2=15\\ \Rightarrow x^2=5\\ \Rightarrow x=\pm\sqrt{5}\)
1. x2 - \(\dfrac{1}{5}.\dfrac{5}{4}=\dfrac{3}{4}\)
<=> x2 - \(\dfrac{1}{4}\) = \(\dfrac{3}{4}\)
<=> x2 = \(\dfrac{3}{4}+\dfrac{1}{4}\)
<=> x2 = 1
<=> x = \(\pm\)1
2. \(\left(\dfrac{1}{2}-x\right)^2=\dfrac{1}{9}\)
<=> \(\dfrac{1}{4}-x+x^2=\dfrac{1}{9}\)
<=> x2 - x = \(\dfrac{1}{9}-\dfrac{1}{4}\)
<=> x2 - x = \(\dfrac{-5}{36}\)
<=> x2 - x - \(\dfrac{-5}{36}\) = 0
Đoạn này dài, mik giải ngoài rồi viết vào nha:
<=> x = \(\dfrac{5}{6}\)
3. x2 - 5 = 5
<=> x2 = 10
<=> x = \(\sqrt{10}\)
4. 3x2 - 1 = 14
<=> 3x2 = 15
<=> x2 = 15 : 3
<=> x2 = 5
<=> x = \(\sqrt{5}\)
a:=>3x=15
=>x=5
b: =>x+3=0,96
=>x=-2,04
c: =>x^2=36
=>x=6 hoặc x=-6
`a, 3/4=(3x)/20`
`3x*4=3*20`
`3x*4=60`
`3x=60 \div 4`
`3x=15`
`x=15 \div 3`
`x=5`
`b, (1,2)/(x+3)=5/4`
`1,2*4=(x+3)*5`
`4,8=(x+3)*5`
`x+3= 4,8 \div 5`
`x+3=0,96`
`x=0,96-3`
`x=-2,04`
`c, (x^2)/32=9/8`
`x^2*8=32*9`
`x^2*8=288`
`x^2=288 \div 8`
`x^2=36`
`x^2=(+-6)^2`
`-> \text {x= 6 hoặc -6}`
\(\left(3-x\right)^3=-\dfrac{27}{64}\)
\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)
\(=>3-x=\dfrac{-3}{4}\)
\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)
\(x=\dfrac{15}{4}\)
________
\(\left(x-5\right)^3=\dfrac{1}{-27}\)
\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)
\(=>x-5=\dfrac{-1}{3}\)
\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)
\(x=\dfrac{14}{3}\)
_____________
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)
\(\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{3}{2}\right)^3\)
\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}+\dfrac{1}{2}\)
\(x=2\)
________
\(\left(2x-1\right)^2=\dfrac{1}{4}\)
\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\) hoặc \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(=>2x-1=\dfrac{1}{2}\) \(2x-1=\dfrac{-1}{2}\)
\(2x=\dfrac{1}{2}+1=\dfrac{1}{2}+\dfrac{2}{2}\) \(2x=\dfrac{-1}{2}+1=\dfrac{-1}{2}+\dfrac{2}{2}\)
\(2x=\dfrac{3}{2}\) \(2x=\dfrac{1}{2}\)
\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\) \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)
\(x=\dfrac{3}{4}\) \(x=\dfrac{1}{4}\)
____________
\(\left(2-3x\right)^2=\dfrac{9}{4}\)
\(\left(2-3x\right)^2=\left(\dfrac{3}{2}\right)^2\) hoặc \(\left(2-3x\right)^2=\left(\dfrac{-3}{2}\right)^2\)
\(=>2-3x=\dfrac{3}{2}\) \(2-3x=\dfrac{-3}{2}\)
\(3x=2-\dfrac{3}{2}=\dfrac{4}{2}-\dfrac{3}{2}\) \(3x=2-\dfrac{-3}{2}=\dfrac{4}{2}+\dfrac{3}{2}\)
\(3x=\dfrac{1}{2}\) \(3x=\dfrac{7}{2}\)
\(x=\dfrac{1}{2}.\dfrac{1}{3}\) \(x=\dfrac{7}{2}.\dfrac{1}{3}\)
\(x=\dfrac{1}{6}\) \(x=\dfrac{7}{6}\)
______________
\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này
(3-x)3=(-\(\dfrac{3}{4}\))3
3-x=-\(\dfrac{3}{4}\)
x=3-(-\(\dfrac{3}{4}\))
x=\(\dfrac{15}{4}\)
a: \(\Leftrightarrow\dfrac{3}{4}:\dfrac{22}{9}-\left|3x-\dfrac{8}{3}\right|=\dfrac{3}{4}\)
\(\Leftrightarrow\left|3x-\dfrac{8}{3}\right|=\dfrac{3}{4}\cdot\dfrac{9}{22}-\dfrac{3}{4}=\dfrac{3}{4}\cdot\dfrac{-13}{22}=\dfrac{-39}{88}\)(loại)
b: \(\Leftrightarrow\left[{}\begin{matrix}3x-2=x-\dfrac{1}{3}\\3x-2=\dfrac{1}{3}-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{5}{3}\\4x=\dfrac{7}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}\\x=\dfrac{7}{12}\end{matrix}\right.\)
\(\dfrac{3x-2}{4}=\dfrac{9}{3x-2}\\ \Rightarrow\left(3x-2\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}3x-2=-6\\3x-2=6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=\dfrac{8}{3}\end{matrix}\right.\)
\(\dfrac{3x-2}{4}=\dfrac{9}{3x-2}\left(đk:x\ne\dfrac{2}{3}\right)\)
\(\Rightarrow\left(3x-2\right)^2=36\)
\(\Rightarrow\left[{}\begin{matrix}3x-2=6\\3x-2=-6\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)