(X-5)^2-3=33
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VV
10 tháng 8 2021
\(2^3x+5^2x=2.\left(5^2+2^3\right)-33\)
\(8x+25x=2.\left(25+8\right)-33\)
\(8x+25x=2.33-33\)
\(8x+25x=66-33\)
\(8x+25x=33\)
\(x+\left(8+25\right)=33\)
\(x+33=33\)
\(x=0\)
VV
10 tháng 8 2021
\(15:\left(x+2\right)=\left(3^3+3\right):10\)
\(15:\left(x+2\right)=\left(27+3\right):10\)
\(15:\left(x+2\right)=30:10\)
\(15:\left(x+2\right)=3\)
\(x+2=15:3\)
\(x+2=5\)
\(x=3\)
12 tháng 3 2021
\(5^{10}:5^8+x=3^{20}:\left(-3\right)^{18}-2^{35}:\left(-2\right)^{33}\\ \Leftrightarrow5^2+x=3^2+\left(-2\right)^2\\ \Leftrightarrow25+x=13\\ \Leftrightarrow x=-12\)
LM
Lê Minh Vũ
CTVHS
27 tháng 7 2023
1) \(2^3\times x-5^2\times x=2\times\left(5^2+2^2\right)-33\)
\(x\times\left(2^3-5^2\right)=2\times\left(25+4\right)-33\)
\(x\times\left(8-25\right)=2\times29-33\)
\(x\times-17=25\)
\(x=-\dfrac{25}{17}\)
2) \(15\div\left(x+2\right)=\left(3^3+3\right)\div1\)
\(15\div\left(x+2\right)=\left(27+3\right)\div1\)
\(15\div\left(x+2\right)=30\div1\)
\(15\div\left(x+2\right)=30\)
\(x+2=\dfrac{1}{2}\)
\(x=-\dfrac{3}{2}\)
3) \(20\div\left(x+1\right)=\left(5^2+1\right)\div13\)
\(20\div\left(x+1\right)=\left(25+1\right)\div13\)
\(20\div\left(x+1\right)=26\div13\)
\(20\div\left(x+1\right)=2\)
\(x+1=20\div2\)
\(x+1=10\)
\(x=9\)
LM
Lê Minh Vũ
CTVHS
27 tháng 7 2023
4) \(320\div\left(x-1\right)=\left(5^3-5^2\right)\div4+15\)
\(320\div\left(x-1\right)=\left(125-25\right)\div4+15\)
\(320\div\left(x-1\right)=100\div4+15\)
\(320\div\left(x-1\right)=25+15\)
\(320\div\left(x-1\right)=40\)
\(x-1=8\)
\(x=9\)
5) \(240\div\left(x-5\right)=2^2\times5^2-20\)
\(240\div\left(x-5\right)=4\times25-20\)
\(240\div\left(x-5\right)=100-20\)
\(240\div\left(x-5\right)=80\)
\(x-5=30\)
\(x=35\)
6) \(70\div\left(x-3\right)=\left(3^4-1\right)\div4-10\)
\(70\div\left(x-3\right)=\left(81-1\right)\div4-10\)
\(70\div\left(x-3\right)=80\div4-10\)
\(70\div\left(x-3\right)=20-10\)
\(70\div\left(x-3\right)=10\)
\(x-3=7\)
\(x=10\)
I
28 tháng 3 2020
đây là đề thi học sinh giỏi Bình định năm 2014-2015 ( mình đc cô giáo cho làm r nên bạn cứ yên tâm là đúng nhá . làm tỷ đề mà zẫn nhớ )
ta có \(x^3=\left(2+\sqrt{3}\right)-\left(2-\sqrt{3}\right)-3\sqrt[3]{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}.x\Rightarrow x^3+3x=2\sqrt{3}\left(1\right)\)
\(y^3=\left(\sqrt{5}+2\right)-\left(\sqrt{5}-2\right)-3\sqrt[3]{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}.y\Rightarrow y^3+3y=4\left(2\right)\)
Trừ theo zế của (1) cho (2) ta được
\(\left(x^3-y^3\right)+3\left(x-y\right)=2\sqrt{3}-4\)
do đó
\(A=\left(x-y\right)^3+3\left(x-y\right)\left(xy+1\right)=x^3-y^3-3\left(x-y\right)xy+3\left(x-y\right)xy+3\left(x-y\right)\)
\(=x^3-y^3+3\left(x-y\right)=2\sqrt{3}-4\)
HH
4
8 tháng 2 2022
1, \(x=-\dfrac{7}{3}-\dfrac{1}{3}=-\dfrac{8}{3}\)
2, \(x=\dfrac{1}{8}-\dfrac{3}{8}=-\dfrac{2}{8}=-\dfrac{1}{4}\)
3, \(x=\dfrac{6}{5}+\dfrac{4}{5}=\dfrac{10}{5}=2\)
4, \(x=\dfrac{7}{3}-\dfrac{2}{3}=\dfrac{5}{3}\)
5, \(x+\dfrac{7}{3}=\dfrac{5}{3}\Leftrightarrow x=\dfrac{5}{3}-\dfrac{7}{3}=-\dfrac{2}{3}\)
8 tháng 2 2022
\(=x+\dfrac{1}{3}=\dfrac{-7}{3}\Leftrightarrow x=\dfrac{-8}{3}\)
\(=\dfrac{1}{8}-x=\dfrac{3}{8}\Leftrightarrow x=\dfrac{-1}{4}\)
\(=x-\dfrac{4}{5}=\dfrac{6}{5}\Leftrightarrow x=2\)
\(=\dfrac{2}{3}+x=\dfrac{7}{3}\Leftrightarrow x=\dfrac{5}{3}\)
\(=x-\dfrac{-7}{3}=\dfrac{5}{3}\Leftrightarrow x=\dfrac{-2}{3}\)
5 tháng 8 2023
`@` `\text {Ans}`
`\downarrow`
`1/2*x + 3/5*(x - 2) = 3`
`=> 1/2*x + 3/5*x - 3/5*2 = 3`
`=> 1/2x + 3/5x - 6/5 = 3`
`=> (1/2 + 3/5)x - 6/5 = 3`
`=> 11/10x - 6/5 = 3`
`=> 11/10x = 3 + 6/5`
`=> 11/10x =21/5`
`=> x = 21/5 \div 11/10`
`=> x = 42/11`
Vậy, `x = 42/11`
____
`3 - (1/6 - x)*2/3 = 2/3`
`=> (1/6 - x)*2/3 = 3 - 2/3`
`=> (1/6 - x)*2/3 = 7/3`
`=> 1/6 - x = 7/3 \div 2/3`
`=> 1/6 - x=7/2`
`=> x = 1/6 - 7/2`
`=> x = -10/3`
Vậy, `x = -10/3.`
5 tháng 8 2023
\(\dfrac{1}{2}\cdot x+\dfrac{3}{5}\left(x-2\right)=3\\ \dfrac{1}{2}\cdot x+\dfrac{3}{5}\cdot x-\dfrac{13}{5}=3\\ \left(\dfrac{1}{2}+\dfrac{3}{5}\right)x-\dfrac{13}{5}=3\\ \dfrac{11}{10}x-\dfrac{13}{5}=3\\ \dfrac{11}{10}x=\dfrac{28}{5}\\ x=\dfrac{28}{5}:\dfrac{11}{10}\\ x=\dfrac{28}{11}\\ 3-\left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=\dfrac{2}{3}\\ \left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=3-\dfrac{2}{3}\\ \left(\dfrac{1}{6}-x\right)\cdot\dfrac{2}{3}=\dfrac{7}{3}\\ \dfrac{1}{6}-x=\dfrac{7}{3}:\dfrac{2}{3}\\ \dfrac{1}{6}-x=\dfrac{7}{2}\\ x=\dfrac{1}{6}-\dfrac{7}{2}\\ x=-\dfrac{10}{3}\)