Giúp em câu 6,7,8,9,10 với ạ
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Bài 9:
6: \(\Leftrightarrow x+\dfrac{4}{3}-\dfrac{12}{5}=\dfrac{5}{2}-\dfrac{1}{4}+\dfrac{7}{8}=\dfrac{20-2+7}{8}=\dfrac{25}{8}\)
=>x=25/8-4/3+12/5=503/120
7: \(\Leftrightarrow\dfrac{1}{7}-\dfrac{1}{5}+x-\dfrac{3}{2}=\dfrac{16}{70}-\dfrac{9}{70}=\dfrac{1}{10}\)
=>x=1/10-1/7+1/5+3/2=58/35
8: \(\Leftrightarrow\dfrac{5}{4}-x+\dfrac{1}{5}-\dfrac{19}{20}=1-\dfrac{1}{2}+\dfrac{7}{10}=\dfrac{6}{5}\)
=>1/2-x=6/5
=>x=1/2-6/5=-7/10
9: \(\Leftrightarrow5-\dfrac{1}{3}+\dfrac{17}{4}-x=1+\dfrac{2-11}{12}=1-\dfrac{9}{12}=\dfrac{1}{4}\)
=>x=14/3+4=26/3
10: \(\Leftrightarrow\dfrac{-1}{12}+x-\dfrac{5}{12}=3-\dfrac{7}{6}+\dfrac{1}{2}\)
=>x-1/2=11/6+1/2
=>x=11/6+1=17/6
\(#nKaizz\)
`9,`
`6, 1 1/3 - 2 2/5 + x= 5/2 - (1/4 - 7/8)`
`4/3 -12/5 + x = 5/2 - (-5/8)`
`4/3 - 12/5 + x=25/8`
`-16/15 + x=25/8`
`x=25/8 - (-16/15)`
`x=25/8 + 16/15 = 503/120`
`7, 1/7 - (1/5 - x + 1 1/2)= 8/35 - 9/70`
`1/7 - (1/5 - x + 3/2)= 1/10`
`1/5 - x + 3/2 = 1/7 -1/10`
`1/5 -x + 3/2 = 3/70`
`1/5 -x=3/70 - 3/2`
`1/5-x=-51/35`
`x=1/5 - (-51/35)`
`x=1/5 + 51/35 = 58/35`
`8, 5/4 - x +(1/5 - 19/20)= 1- (1/2 - 7/10)`
`5/4 - x+ -3/4 = 1- (-1/5)`
`5/4 - x+ -3/4 = 6/5`
`5/4 - x=6/5 - (-3/4)`
`5/4 - x=39/20`
`x=5/4 - 39/20`
`x=-7/10`
`9, 5- (1/3 - 4 1/4+x) = 1+ (1/6 - 11/12)`
`5- (1/3- 4 1/4+x)=1+ -3/4 `
`5- (1/3 - 4 1/4+x) = 1/4`
`1/3- 4 1/4 +x=5-1/4`
`-47/12 +x=19/4`
`x=19/4 - (-47/12)`
`x=26/3`
`10, 1/4 - (1/3-x+5/12)=3-(1 1/6 - 1/2)`
`1/4-(1/3-x+5/12)=3-2/3`
`1/4 - (1/3-x+5/12) =7/3`
`1/3 - x+5/12 = 1/4 - 7/3`
`1/3-x+5/12 = -25/12`
`1/3-x = (-25/12)-5/12`
`1/3-x = -5/2`
`x=1/3 - (-5/2)`
`x=17/6`
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b: Tọa độ là:
\(\left\{{}\begin{matrix}\dfrac{2}{3}x=-\dfrac{1}{3}x+2\\y=\dfrac{2}{3}x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{4}{3}\end{matrix}\right.\)
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Câu 7:
a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
b, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{10}.100\%=56\%\\\%m_{CuO}=44\%\end{matrix}\right.\)
c, \(n_{CuO}=\dfrac{10-0,1.56}{80}=0,055\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{Fe}+n_{CuO}=0,155\left(mol\right)\)
\(\Rightarrow C\%_{H_2SO_4}=\dfrac{0,155.98}{100}.100\%=15,19\%\)
d, Theo PT: \(\left\{{}\begin{matrix}n_{FeSO_4}=n_{Fe}=0,1\left(mol\right)\\n_{CuSO_4}=n_{CuO}=0,055\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,1.152=15,2\left(g\right)\\m_{CuSO_4}=0,055.160=8,8\left(g\right)\end{matrix}\right.\)
Câu 8:
a, \(CuCO_3+2HCl\rightarrow CuCl_2+CO_2+H_2O\)
b, \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{CuCO_3}=n_{CO_2}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuCO_3}=\dfrac{0,15.124}{20}.100\%=93\%\\\%m_{CuCl_2}=7\%\end{matrix}\right.\)
c, \(n_{HCl}=2n_{CO_2}=0,3\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)
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Para 1 - b
Para 2 - a
Para 3 - c
T - F - T - T - NG
1 B
2 A
3 D
4 D
5 A
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a) Xét ΔMNI vuông tại M và ΔHPI vuông tại P có
\(\widehat{MIN}=\widehat{HIP}\)(hai góc đối đỉnh)
Do đó: ΔMNI\(\sim\)ΔHPI(g-g)
b) Ta có: ΔMNI\(\sim\)ΔHPI(cmt)
nên \(\widehat{MNI}=\widehat{HPI}\)(hai góc tương ứng)
hay \(\widehat{MNI}=\widehat{MPK}\)
Xét ΔMNI vuông tại M và ΔMPK vuông tại M có
\(\widehat{MNI}=\widehat{MPK}\)(cmt)
Do đó: ΔMNI\(\sim\)ΔMPK(g-g)
Suy ra: \(\dfrac{MN}{MP}=\dfrac{MI}{MK}\)(Các cặp cạnh tương ứng tỉ lệ)
hay \(\dfrac{MN}{MI}=\dfrac{MP}{MK}\)
Xét ΔMNP vuông tại M và ΔMIK vuông tại M có
\(\dfrac{MN}{MI}=\dfrac{MP}{MK}\)(cmt)
Do đó: ΔMNP\(\sim\)ΔMIK(c-g-c)
6) \(=3\left(x-y\right)\left(1-2a\right)\)
7) \(=\left(x+2\right)\left(5+8y\right)\)
8) \(=3x\left(x-1\right)+6y\left(x-1\right)=3\left(x-1\right)\left(x+2y\right)\)
9) \(=2\left(2x-1\right)\left(2x+3\right)\)
10) \(=3x\left(x+3\right)\left(x-2\right)\)
1. 4x2 - 4x
= 4x(x - 1)
2. 3ax + 12x - 9
= 3(ax + 4x - 3)
3. 6xy + 2xy2 - 8x2
= 2x(3y + y2 - 4x)
4. -4ay + 10a - 8a2y
= -2a(2y - 5 + 8ay)
5. 10x3 - 8ax2
= 2x2(5x - 4a)
6. 3(x - y) - 6a(x - y)
= (3 - 6a)(x - y)
= 3(1 - 2a)(x - y)
7. 5(x + 2) + 8y(x + 2)
= (5 + 8y)(x + 2)
8. 3x(x - 1) - 6y(1 - x)
= 3x(x - 1) + 6y(x - 1)
= (3x + 6y)(x - 1)
= 3(x + 2y)(x - 1)
9. 4x(2x - 1) + 6(2x - 1)
= (4x + 6)(2x - 1)
= 2(2x + 3)(2x - 1)
10. 3x2(x + 3) - 6x(3 + x)
= 3x2(x + 3) - 6x(x + 3)
= (3x2 - 6x)(x + 3)
= 3x(x - 2)(x + 3)