Nếu \(\dfrac{x+4}{18}=\dfrac{1}{3}\) thì x có giá trị bằng bao nhiu
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`B17:`
`a)` Với `x \ne +-3` có:
`A=[x+15]/[x^2-9]+2/[x+3]`
`A=[x+15+2(x-3)]/[(x-3)(x+3)]`
`A=[x+15+2x-6]/[(x-3)(x+3)]`
`A=[3x+9]/[(x-3)(x+3)]=3/[x-3]`
`b)A=[-1]/2<=>3/[x-3]=-1/2<=>-x+3=6<=>x=-3` (ko t/m)
`=>` Ko có gtr nào của `x` t/m
`c)A in ZZ<=>3/[x-3] in ZZ`
`=>x-3 in Ư_3`
Mà `Ư_3={+-1;+-3}`
`@x-3=1=>x=4`
`@x-3=-1=>x=2`
`@x-3=3=>x=6`
`@x-3=-3=>x=0`
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`B18:`
`a)M=1/3` `ĐK: x \ne +-4`
`<=>(4/[x-4]-4/[x+4]).[x^2+8x+16]/32=1/3`
`<=>[4(x+4)-4(x-4)]/[(x-4)(x+4)].[(x+4)^2]/32=1/3`
`<=>32/[x-4].[x+4]/32=1/3`
`<=>3x+12=x-4`
`<=>x=-8` (t/m)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,ĐK:x\ne\pm2\\ b,A=\dfrac{x^2+4x+4+x^2-4x+4+16}{2\left(x-2\right)\left(x+2\right)}\\ A=\dfrac{2x^2+32}{2\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+16}{x^2-4}\\ c,A=-3\Leftrightarrow-3x^2+12=x^2+16\\ \Leftrightarrow4x^2=-4\Leftrightarrow x\in\varnothing\)
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a: \(A=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\left(\dfrac{\sqrt{3}-1}{4}\right)^2}=\dfrac{\sqrt{12+2\sqrt{3}}}{4}\)
\(\Rightarrow2\cos\alpha=\dfrac{\sqrt{12+2\sqrt{3}}}{2}\). Chọn B.
![](https://rs.olm.vn/images/avt/0.png?1311)
P = (\(\dfrac{1}{\sqrt{x}-1}\) - \(\dfrac{1}{\sqrt{x}}\)) : (\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\) - \(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)) với 0 < \(x\) ≠ 1; 4
P = \(\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}\): (\(\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right).\left(\sqrt{x-2}\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}\))
P = \(\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\): \(\dfrac{x-1-\left(x-4\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}\)
P = \(\dfrac{1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\) : \(\dfrac{3}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}\)
P = \(\dfrac{1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\) \(\times\) \(\dfrac{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}{3}\)
P = \(\dfrac{\sqrt{x}-2}{3.\sqrt{x}}\)
P = \(\dfrac{\sqrt{x}.\left(\sqrt{x}-2\right)}{3x}\)
b, P = \(\dfrac{1}{4}\)
⇒ \(\dfrac{\sqrt{x}.\left(\sqrt{x}-2\right)}{3x}\) = \(\dfrac{1}{4}\)
⇒4\(x\) - 8\(\sqrt{x}\) = 3\(x\)
⇒ 4\(x\) - 8\(\sqrt{x}\) - 3\(x\) = 0
\(x\) - 8\(\sqrt{x}\) = 0
\(\sqrt{x}\).(\(\sqrt{x}\) - 8) = 0
\(\left[{}\begin{matrix}x=0\\\sqrt{x}=8\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=64\end{matrix}\right.\)
\(x=0\) (loại)
\(x\) = 64
![](https://rs.olm.vn/images/avt/0.png?1311)
A = \(\left(\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+2}\right)\cdot\dfrac{4x-4}{5}\) (ĐK: x \(\ge\) 0; x \(\ne\) 1)
A = \(\left(\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)
A = \(\left(\dfrac{\left(\sqrt{x}+1\right)^2}{2\left(x-1\right)}+\dfrac{6}{2\left(x-1\right)}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{2\left(x-1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)
A = \(\left(\dfrac{x+2\sqrt{x}+1+6-x-3\sqrt{x}+\sqrt{x}+3}{2\left(x-1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)
A = \(\dfrac{10}{2\left(x-1\right)}\cdot\dfrac{4\left(x-1\right)}{5}\)
A = 4
Vậy A không phụ thuộc vào x
Chúc bn học tốt!
Ta có: \(A=\left(\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+2}\right)\cdot\dfrac{4x-4}{5}\)
\(=\dfrac{x+2\sqrt{x}+1+6-\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{5}\)
\(=\dfrac{x+2\sqrt{x}+7-x-2\sqrt{x}+3}{1}\cdot\dfrac{2}{5}\)
\(=10\cdot\dfrac{2}{5}=4\)
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\(PT\Leftrightarrow\dfrac{5}{2}\sqrt{2x+1}-\sqrt{\dfrac{\dfrac{2x+1}{2}}{2}}=\dfrac{3}{2}\\ \Leftrightarrow\dfrac{5}{2}\sqrt{2x+1}-\dfrac{1}{2}\sqrt{2x+1}=\dfrac{3}{2}\\ \Leftrightarrow2\sqrt{2x+1}=\dfrac{3}{2}\\ \Leftrightarrow\sqrt{2x+1}=\dfrac{3}{4}\\ \Leftrightarrow2x+1=\dfrac{9}{16}\\ \Leftrightarrow2x=-\dfrac{7}{16}\\ \Leftrightarrow x=-\dfrac{7}{32}\\ \Leftrightarrow a=-\dfrac{7}{32}\\ \Leftrightarrow1-36a=1+36\cdot\dfrac{7}{32}=...\)
trình bày đày đủ giúp em ạ
\(\Rightarrow\left(x+4\right)\times3=18\\ \Leftrightarrow x+4=\dfrac{18}{3}=6\\ \Leftrightarrow x=6-4=2\)