x+25%x=1 3/8
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1: =>x=8-35=-27
2: =>15-4+x=6
=>x+11=6
hay x=-5
3: =>-30+25-x=-1
=>x+5=1
hay x=-4
4: =>x-(-13)=-8
=>x+13=-8
hay x=-21
5: =>x-29-17+38=-9
=>x-8=-9
hay x=-1
G(x) = 8(x + 1)³ + 1
G(x) = 0
⇒ 8(x + 1)³ + 1 = 0
8(x + 1)³ = -1
(x + 1)³ = -1/8
(x + 1)³ = (-1/2)³
x + 1 = -1/2
x = -1/2 - 1
x = -3/2
Vậy nghiệm của G(x) là x = -3/2
H(x) = 8/9 - 2((x - 1)²
H(x) = 0
⇒ 8/9 - 2(x - 1)² = 0
2(x - 1)² = 8/9
(x - 1)² = 8/9 : 2
(x - 1)² = 4/9
x - 1 = 2/9 hoặc x - 1 = -2/9
*) x - 1 = 2/9
x = 2/9 + 1
x = 11/9
*) x - 1 = -2/9
x = -2/9 + 1
x = 7/9
Vậy nghiệm của H(x) là x = 7/9; x = 11/9
a) \(7x\left(x+1\right)-3\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(7x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\7x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{7}\end{matrix}\right.\)
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 => \(\left[{}\begin{matrix}x+8=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=3\end{matrix}\right.\)
c) \(x^2-10x=-25\Rightarrow x^2-10x+25=0\Rightarrow\left(x-5\right)^2=0\Rightarrow x=5\)
d) Giống câu c
a)
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 =>
c)
1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)
\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)
\(=\dfrac{-1}{2}+\dfrac{4}{5}\)
\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)
2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)
\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)
\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)
3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)
\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)
\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{17}{7}\)
4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)
\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)
\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)
\(x^3+1\)
\(=x^3+1^3\)
\(=\left(x+1\right)\left(x^2-x+1\right)\)
______
\(8+x^3\)
\(=2^3+x^3\)
\(=\left(2+x\right)\left(4-2x+x^2\right)\)
______
\(27x^3-64y^3\)
\(=\left(3x\right)^3-\left(4y\right)^3\)
\(=\left(3x-4y\right)\left(9x^2+12xy+16y^2\right)\)
______
\(\dfrac{x^3}{64}-\dfrac{1}{125}\)
\(=\left(\dfrac{x}{4}\right)^3-\left(\dfrac{1}{5}\right)^3\)
\(=\left(\dfrac{x}{4}-\dfrac{1}{5}\right)\left(\dfrac{x^2}{16}+\dfrac{x}{20}+\dfrac{1}{25}\right)\)
x^3+1=(x+1)(x^2-x+1)
x^3+8=(x+2)(x^2-2x+4)
27x^3-64y^3=(3x-4y)(9x^2+12xy+16y^2)
\(\dfrac{x^3}{64}-\dfrac{1}{25}=\left(\dfrac{1}{4}x-\sqrt[3]{\dfrac{1}{5}}\right)\left(\dfrac{1}{16}x^2+\dfrac{1}{4\sqrt[3]{5}}\cdot x+\dfrac{1}{\sqrt[3]{25}}\right)\)
\(2\times3\times5+25\times8\times4+70+4\times2\times25\)
\(=15\times2+100\times8+70+100\times2\)
\(=30+800+70+200\)
\(=\left(30+70\right)+\left(800+200\right)\)
\(=100+100\times10\)
\(=100+1000\)
\(=1100\)
`x + 25%x = 1 3/8`
`=> 5/4x = 11/8`
`=> 10/8x = 11/8`
`=> x = 11/8:10/8`
`=> x = 11/10`
Vậy `...`
\(x+25\%x=1\dfrac{3}{8}\)
\(x+\dfrac{1}{4}x=\dfrac{11}{8}\)
\(x.\left(\dfrac{1}{4}+1\right)=\dfrac{11}{8}\)
\(x.\dfrac{5}{4}=\dfrac{11}{8}\)
\(x=\dfrac{11}{8}:\dfrac{5}{8}\)
\(x=\dfrac{11}{8}\times\dfrac{8}{5}\)
\(x=\dfrac{11}{10}\)
\(Vẫy=\dfrac{11}{10}\)