giá trị của \(x\) thỏa mãn \(\dfrac{x-2}{0,2}=\dfrac{0,7}{4}\)
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\(\dfrac{15}{6-x}-\dfrac{3}{6-x}=4\left(x\ne6\right)\\ =>15-3=24-4x\\ < =>4x=24-15+3\\ < =>4x=12\\ < =>x=3\left(tmđk\right)\)
a) A = \(\dfrac{1}{x-1}-\dfrac{4}{x+1}+\dfrac{8x}{\left(x-1\right)\left(x+1\right)}\)
= \(\dfrac{x+1-4x+4+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{5x+5}{\left(x-1\right)\left(x+1\right)}=\dfrac{5}{x-1}\) => đpcm
b) \(\left|x-2\right|=3=>\left[{}\begin{matrix}x-2=3< =>x=5\left(C\right)\\x-2=-3< =>x=-1\left(L\right)\end{matrix}\right.\)
Thay x = 5 vào A, ta có:
A = \(\dfrac{5}{5-1}=\dfrac{5}{4}\)
c) Để A nguyên <=> \(5⋮x-1\)
x-1 | -5 | -1 | 1 | 5 |
x | -4(C) | 0(C) | 2(C) | 6(C) |
\(3,=\left(\dfrac{13}{25}-\dfrac{38}{25}\right)+\left(\dfrac{14}{9}-\dfrac{5}{9}\right)=-1+1=0\\ 4,=\left(\dfrac{4}{9}\right)^5\cdot\left(\dfrac{9}{49}\right)^5=\left(\dfrac{4}{9}\cdot\dfrac{9}{49}\right)^5=\left(\dfrac{4}{49}\right)^5\\ 5,\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x-y}{5-3}=\dfrac{x+y}{5+3}=\dfrac{2}{2}=\dfrac{x+y}{8}\Rightarrow x+y=8\\ 6,\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\Rightarrow2\text{ giá trị}\\ 7,=\dfrac{3^{10}\cdot2^{30}}{2^9\cdot3^9\cdot2^{20}}=2\cdot3=6\)
`a/(x+1)+b/(x-2)=(a(x-2)+b(x+1))/((x+1)(x-2))`
`=(ax-2a+bx+b)/(x^2-x-2)`
`=((a+b)x+(-2a+b))/(x^2-x-2)`
``
Theo đề bài: `((a+b)x+(-2a+b))/(x^2-x-2)=(32x-19)/(x^2-x-2)`
Đồng nhất hệ số ta được: `{(a+b=32),(-2a+b=-19):}`
`<=>{(a+b=32),(2a-b=19):}`
`<=>{(3a=51),(a+b=32):}`
`<=>{(a=17),(17+b=32):}`
`<=>{(a=17),(b=15):}`
2,035 ?
\(\dfrac{x-2}{0,2}=\dfrac{0,7}{4}\)
\(\Rightarrow4\left(x-2\right)=0,7\cdot0,2\)
\(\Rightarrow4x-8=0,14\)
\(\Rightarrow4x=0,14+8\)
\(\Rightarrow4x=8,14\)
\(\Rightarrow x=8,14:4\)
\(\Rightarrow x=2,035\)