giải giúp mình câu C với câu D với ạ
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c: ĐKXĐ: x<>8
\(\dfrac{3}{2x-16}+\dfrac{3x-20}{x-8}+\dfrac{1}{8}=\dfrac{13x-102}{3x-24}\)
=>\(\dfrac{9}{6\left(x-8\right)}+\dfrac{18x-120}{6\left(x-8\right)}-\dfrac{26x-204}{6\left(x-8\right)}=\dfrac{-1}{8}\)
=>\(\dfrac{18x-111-26x+204}{6\left(x-8\right)}=\dfrac{-1}{8}\)
=>\(\dfrac{-8x+93}{6x-48}=\dfrac{-1}{8}\)
=>\(\dfrac{8x-93}{6x-48}=\dfrac{1}{8}\)
=>8(8x-93)=6x-48
=>64x-744-6x+48=0
=>58x=696
=>x=12
d: ĐKXĐ: x<>1; x<>-1
\(\dfrac{6}{x^2-1}+5=\dfrac{8x-1}{4x+4}+\dfrac{12x-1}{4x-4}\)
=>\(\dfrac{24}{4\left(x-1\right)\left(x+1\right)}+\dfrac{20\left(x^2-1\right)}{4\left(x-1\right)\left(x+1\right)}=\dfrac{\left(8x-1\right)\left(x-1\right)+\left(12x-1\right)\left(x+1\right)}{4\left(x-1\right)\left(x+1\right)}\)
=>8x^2-9x+1+12x^2+12x-x-1=24+20x^2-20
=>20x^2+2x=20x^2+4
=>2x=4
=>x=2(loại)
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\(b,N=\left(2x-1\right)^2-4\ge-4\\ N_{min}=-4\Leftrightarrow x=\dfrac{1}{2}\\ c,P=\left(2x-5\right)^2+6\left(2x-5\right)+9-4\\ P=\left(2x-5+3\right)^2-4=\left(2x-2\right)^2-4\ge-4\\ P_{min}=-4\Leftrightarrow x=1\\ d,Q=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\\ Q=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\\ Q_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
6a.
$M=x^2-x+1=(x^2-x+\frac{1}{4})+\frac{3}{4}$
$=(x-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}$
Vậy $M_{\min}=\frac{3}{4}$ khi $x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}$
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a) Gọi x, y lần lượt là số mol Al, Fe
2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Fe + H2SO4 → FeSO4+ H2
\(\left\{{}\begin{matrix}27x+56y=5,54\\\dfrac{3}{2}x+y=\dfrac{3,584}{22,4}\end{matrix}\right.\)
=> x=0,06 , y =0,07
=> \(m_{Al}=1,62\left(g\right);m_{Fe}=3,92\left(g\right)\)
b) \(n_{H_2SO_4\left(pứ\right)}=n_{H_2}=0,16\left(mol\right)\)
=> \(m_{H_2SO_4\left(pứ\right)}=0,16.98=15,68\left(g\right)\)
c) \(m_{ddH_2SO_4}=\dfrac{15,68}{20\%}=78,4\left(g\right)\)
c) 2NaOH + H2SO4 → Na2SO4 + 2H2O
\(n_{H_2SO_4\left(dư\right)}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}.0,25.0,6=0,075\left(mol\right)\)
=> \(m_{H_2SO_4\left(bđ\right)}=15,68+0,075.98=23,03\left(g\right)\)
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Câu 5
2Al+3H2SO4->Al2(SO4)3+3H2
0,2------------------0,1----------0,3 mol
`n_(H_2)=(6,72)/(22,4)=0,3 mol`
`=>m_(Al_2(SO_4)_3)=0,1.342=34,2g`
c)`3H_2+Fe_2O_3->2Fe+3H_2O`to
0,3----------------------0,2 mol
H=95%
`=>m_(Fe)=0,2.56.95%=10,64g`
`#YBTran:3`
a) \(A=\dfrac{\sqrt[]{x}+2}{\sqrt[]{x}-5}\) có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt[]{x}-5\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt[]{x}\ne5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne25\end{matrix}\right.\)
Khi \(x=16\Rightarrow A=\dfrac{\sqrt[]{16}+2}{\sqrt[]{16}-5}=\dfrac{4+2}{4-5}=-6\)
b) \(B=\dfrac{3}{\sqrt[]{x}+5}+\dfrac{20-2\sqrt[]{x}}{x-25}\)
B có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x-25\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne25\end{matrix}\right.\)
\(\Leftrightarrow B=\dfrac{3\left(\sqrt[]{x}-5\right)+20-2\sqrt[]{x}}{\left(\sqrt[]{x}+5\right)\left(\sqrt[]{x}-5\right)}\)
\(\Leftrightarrow B=\dfrac{3\sqrt[]{x}-15+20-2\sqrt[]{x}}{\left(\sqrt[]{x}+5\right)\left(\sqrt[]{x}-5\right)}\)
\(\Leftrightarrow B=\dfrac{\sqrt[]{x}+5}{\left(\sqrt[]{x}+5\right)\left(\sqrt[]{x}-5\right)}\)
\(\Leftrightarrow B=\dfrac{1}{\sqrt[]{x}-5}\left(dpcm\right)\)
c) \(A=\dfrac{\sqrt[]{x}+2}{\sqrt[]{x}-5}\in Z\left(x\in Z\right)\)
\(\Leftrightarrow\sqrt[]{x}+2⋮\sqrt[]{x}-5\)
\(\Leftrightarrow\sqrt[]{x}+2-\left(\sqrt[]{x}-5\right)⋮\sqrt[]{x}-5\)
\(\Leftrightarrow\sqrt[]{x}+2-\sqrt[]{x}+5⋮\sqrt[]{x}-5\)
\(\Leftrightarrow7⋮\sqrt[]{x}-5\)
\(\Leftrightarrow\sqrt[]{x}-5\in U\left(7\right)=\left\{-1;1;-7;7\right\}\)
\(\Leftrightarrow x\in\left\{16;36;144\right\}\)
d) \(A>B\left(2\sqrt[]{x}+5\right)\)
\(\Leftrightarrow\dfrac{\sqrt[]{x}+2}{\sqrt[]{x}-5}>\dfrac{1}{\sqrt[]{x}-5}\left(2\sqrt[]{x}+5\right)\)
\(\Leftrightarrow\sqrt[]{x}+2>2\sqrt[]{x}+5\)
\(\Leftrightarrow\sqrt[]{x}< -3\)
mà \(\sqrt[]{x}\ge0\)
\(\Leftrightarrow x\in\varnothing\)