(x-3)^5=(x-3) ^3
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a: =>x-2/5=3/4:1/3=3/4*3=9/4
=>x=9/4+2/5=45/20+8/20=53/20
b: =>x-2/3=7/3:4/5=7/3*5/4=35/12
=>x=35/12+2/3=43/12
c: 1/3(x-2/5)=4/5
=>x-2/5=4/5*3=12/5
=>x=12/5+2/5=14/5
d: =>2/3x-1/3-1/4x+1/10=7/3
=>5/12x-7/30=7/3
=>5/12x=7/3+7/30=77/30
=>x=77/30:5/12=154/25
e: \(\Leftrightarrow x\cdot\dfrac{3}{7}-\dfrac{2}{7}+\dfrac{1}{2}-\dfrac{5}{4}x+\dfrac{5}{2}=0\)
=>\(x\cdot\dfrac{-23}{28}=\dfrac{2}{7}-3=\dfrac{-19}{7}\)
=>x=19/7:23/28=76/23
f: =>1/2x-3/2+1/3x-4/3+1/4x-5/4=1/5
=>13/12x=1/5+3/2+4/3+5/4=257/60
=>x=257/65
i: =>x^2-2/5x-x^2-2x+11/4=4/3
=>-12/5x=4/3-11/4=-17/12
=>x=17/12:12/5=85/144
Bạn cần viết đề bài bằng công thức toán để được hỗ trợ tốt hơn.
a) \(\left(3x-1\right).\left(\frac{-1}{2}x+5\right)=0\)
\(\Rightarrow3x-1=0\Rightarrow3x=1\Rightarrow x=\frac{1}{3}\)
\(\frac{-1}{2}x+5=0\Rightarrow\frac{-1}{2}x=-5\Rightarrow x=10\)
b) \(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=x+\frac{1}{5}\)
\(3x-\frac{3}{2}-5x-3=x+\frac{1}{5}\)
\(\Rightarrow3x-5x-x=\frac{1}{5}+\frac{3}{2}+3\)
\(-3x=\frac{47}{10}\)
\(x=\frac{-47}{30}\)
c) \(-5.\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(-5x-1-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(-5x-\frac{1}{2}x-\frac{3}{2}x=\frac{-5}{6}+1-\frac{1}{3}\)
\(-7x=\frac{-1}{6}\)
\(x=\frac{1}{42}\)
d) \(3.\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(3.\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\)
\(\left(3x-\frac{1}{2}\right)^3=\frac{-1}{27}\)
\(\left(3x-\frac{1}{2}\right)^3=\left(\frac{-1}{3}\right)^3\)
\(\Rightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(3x=\frac{1}{6}\)
\(x=\frac{1}{18}\)
Học tốt nhé bn!
a) \(\left(3^2-2\right).\left(x-12+35\right)\)\(=\)\(5^2+279:5\)
\(7.\left(x-12+35\right)=80,8\)
\(x-12+35=80,8:7\)
\(x-12+35=\frac{404}{35}\)
\(x-12=\frac{404}{35}-35\)
\(x-12=\frac{-821}{35}\)
\(x=\frac{-821}{35}+12\)
\(x=\frac{-401}{35}\)
b) \(260:\left(x+4\right)\)\(=\)\(5\left(2^3+5\right)-3\left(3^2+2^2\right)\)
\(260:\left(x+4\right)=26\)
\(x+4=260:26\)
\(x+4=10\)
\(x=10-4\)
\(x=6\)
c) \(7^{x-3}=343\)
\(7^x:7^3=343\)
\(7^x=343.7^3\)
\(7^x=117649\)
vì \(117649=7^6\Rightarrow x=6\)
d) \(\left(x-3\right)^{2017}=\left(x-3\right)^{2016}\)
\(\Rightarrow\hept{\begin{cases}x-3=1\\x-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=4\\x=3\end{cases}}}\)
e) \(\left(2^3+3\right).\left(x-5\right)+14\)\(=\)\(5^2+124:2^2\)
\(\left(2^3+3\right).\left(x-5\right)+14=56\)
\(\left(2^3+3\right).\left(x-5\right)=56-14\)
\(\left(2^3+3\right).\left(x-5\right)=42\)
\(x-5=42:\left(2^3+3\right)\)
\(x-5=\frac{42}{11}\)
\(x=\frac{42}{11}+5\)
\(x=\frac{97}{11}\)
a) 2 x 2 = 4 3 x 3 = 9
2 x 4 = 8 3 x 5 = 15
2 x 6 = 12 3 x 7 = 21
2 x 8 = 16 3 x 9 = 27
4 x 4 = 16 5 x 5 = 25
4 x 2 = 8 5 x 7 = 35
4 x 6 = 24 5 x 9 = 45
4 x 8 = 32 5 x 3 = 15
b) 200 x 4 = 800 300 x 2 = 600
200 x 2 = 400 300 x 3 = 900
400 x 2 = 800 500 x 1 = 500
100 x 4 = 400 100 x 3 = 300
`@` `\text {Ans}`
`\downarrow`
`1)`
\(x+\dfrac{1}{2}=\dfrac{5}{3}\)
`\Rightarrow` \(x=\dfrac{5}{3}-\dfrac{1}{2}\)
`\Rightarrow`\(x=\dfrac{7}{6}\)
Vậy, `x =`\(\dfrac{7}{6}\)
`2)`
\(\dfrac{3}{5}-x=\dfrac{1}{3}\)
`\Rightarrow`\(x=\dfrac{3}{5}-\dfrac{1}{3}\)
`\Rightarrow`\(x=\dfrac{4}{15}\)
Vậy, `x =`\(\dfrac{4}{15}\)
`3)`
\(\dfrac{3}{4}+x=\dfrac{7}{2}\)
`\Rightarrow`\(x=\dfrac{7}{2}-\dfrac{3}{4}\)
`\Rightarrow`\(x=\dfrac{11}{4}\)
Vậy, \(x=\dfrac{11}{4}\)
`4)`
\(x-\dfrac{4}{3}=\dfrac{7}{9}\)
`\Rightarrow`\(x=\dfrac{7}{9}+\dfrac{4}{3}\)
`\Rightarrow`\(x=\dfrac{19}{9}\)
Vậy, `x=`\(\dfrac{19}{9}\)
`5)`
\(x-\dfrac{5}{6}=\dfrac{7}{3}\)
`\Rightarrow`\(x=\dfrac{7}{3}+\dfrac{5}{6}\)
`\Rightarrow x =`\(\dfrac{19}{6}\)
Vậy, `x=`\(\dfrac{19}{6}\)
`6)`
\(x-\dfrac{1}{5}=\dfrac{9}{10}\)
`\Rightarrow x=`\(\dfrac{9}{10}+\dfrac{1}{5}\)
`\Rightarrow x=`\(\dfrac{11}{10}\)
Vậy, `x=`\(\dfrac{11}{10}\)
1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)
\(=x^2+4x+4+x^2-6x+9\)
\(=2x^2-2x+13\)
2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)
\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)
\(=-2x+7\)
3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)
\(=x^2-25-x^2-10x-25\)
=-10x-50
4) Ta có: \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)
\(=x^2-6x+9-x^2+16\)
=-6x+25
5) Ta có: \(\left(y^2-6y+9\right)-\left(y-3\right)^2\)
\(=y^2-6y+9-y^2+6y-9\)
=0
6) Ta có: \(\left(2x+3\right)^2-\left(2x-3\right)\left(2x+3\right)\)
\(=4x^2+12x+9-4x^2+9\)
=12x+18