Mn giải giúp mik vs 
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DT
0
22 tháng 11 2021
\(n_{CuSO_4}=0,2x\left(mol\right)\)
\(Fe+CuSO_4\rightarrow FeSO_4+Cu\)
0,2x 0,2x 0,2x
\(m_{tăng}=m_{Cu}-m_{Fe}=64\cdot0,2x-56\cdot0,2x=1,6\)
\(\Rightarrow x=1M\)
Chọn C.
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
22 tháng 11 2021
\(n_{Fe}=n_{FeSO_4}=n_{Cu}=n_{CuSO_4}=0,2.x\left(mol\right)\\ Fe+CuSO_4\rightarrow FeSO_4+Cu\\ m_{t\text{ăn}g}=m_{Cu.b\text{á}m.v\text{ào}}-m_{Fe.tan.ra}\\ \Leftrightarrow1,6=64.0,2x-56.0,2x\\ \Leftrightarrow x=1\\ \Rightarrow C\)
Giải pt . Mn giúp mik vs
4 tháng 8 2023
ĐK: \(-1\le x\le1\)
Đặt \(\sqrt{1-x}=a;\sqrt{x+1}=b\Rightarrow3-x=2a^2+b^2\)
\(pt\Leftrightarrow2a-b+3ab=2a^2+b^2\)
\(\Leftrightarrow2a^2+b^2-2a+b-3ab=0\)
\(\Leftrightarrow2a^2-a\left(3b+2\right)+b^2+b=0\)
\(\Delta=\left(3b+2\right)^2-4.2.\left(b^2+b\right)=9b^2+12b+4-8b^2-8b\)
\(=b^2+4b+4=\left(b+2\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}a=\dfrac{3b+2-\left(b+2\right)}{4}=\dfrac{2b}{4}=\dfrac{b}{2}\Leftrightarrow2a=b\left(1\right)\\a=\dfrac{3b+2+b+2}{4}=\dfrac{4b+4}{4}=b+1\left(2\right)\end{matrix}\right.\)
pt (1) \(\Leftrightarrow2\sqrt{1-x}=\sqrt{x+1}\)
\(\Leftrightarrow4\left(1-x\right)=x+1\)
\(\Leftrightarrow5x=3\Leftrightarrow x=\dfrac{5}{3}\left(tm\right)\)
\(pt\left(2\right)\Leftrightarrow\sqrt{1-x}=1+\sqrt{x+1}\)
\(\Leftrightarrow1-x=1+x+1+2\sqrt{x+1}\)
\(\Leftrightarrow-1-2x=2\sqrt{x+1}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{1}{2}\\4x^2+4x+1=4x+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{1}{2}\\4x^2=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{1}{2}\\\left[{}\begin{matrix}x=\dfrac{\sqrt{3}}{2}\left(l\right)\\x=-\dfrac{\sqrt{3}}{2}\left(tm\right)\end{matrix}\right.\end{matrix}\right.\)
Vậy, pt có tập nghiệm là: \(S=\left\{-\dfrac{\sqrt{3}}{2};\dfrac{5}{3}\right\}\)
1 tháng 7 2023
`a,` ĐKXĐ: `x>=0;x\ne1`
`A=...=(sqrtx(1+sqrtx)+sqrtx(1-sqrtx)+sqrtx-3)/((1-sqrtx)(1+sqrtx))`
`=(sqrtx+x+sqrtx-x+sqrtx-3)/((1-sqrtx)(1+sqrtx))`
`=(3sqrtx-3)/((1-sqrtx)(1+sqrtx))`
`=-3/(1+sqrtx)`
`b,A=-3/(1+sqrtx)`
Vì `x>=0` nên `1+sqrtx>=1` nên `3/(1+sqrtx)<=3` suy ra `A>=-3`
Dấu "=" xảy ra `<=>x=0`
Vậy `A_(min)=-3<=>x=0`
SV
Sinh Viên NEU
CTVVIP
31 tháng 12 2023
13 I wish I could help her with her business
14 The children said that they were waiting for the school bus
15 The teacher asked his students to listen to her and not to make a noise
16 My sister studied hard so she completed her exam successfully
17 She didn't decide what to wear to the party
18 She was tired so she went home yesterday
19 He has studied English for four years
Câu 1 :
\(a,5\left(x+2\right)=2\left(x-4\right)\)
\(\Leftrightarrow5x+10=2x-8\)
\(\Leftrightarrow5x-2x=-8-10\)
\(\Leftrightarrow3x=-18\)
\(\Leftrightarrow x=-6\)
\(b,x\left(x+2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{3;2\right\}\)
\(c,\dfrac{2x-5}{4}-\dfrac{x+1}{3}=\dfrac{1}{2}\)
\(\Leftrightarrow3\left(2x-5\right)-4\left(x+1\right)=6\)
\(\Leftrightarrow6x-15-4x-4=6\)
\(\Leftrightarrow6x-4x=6+4+15\)
\(\Leftrightarrow2x=25\)
\(\Leftrightarrow x=\dfrac{25}{2}\)
Vậy \(S=\left\{\dfrac{25}{2}\right\}\)
\(d,\dfrac{3}{x-2}-\dfrac{6}{x+2}=\dfrac{-x}{x^2-4}\left(đkxđ:x\ne\pm2\right)\)
\(\Leftrightarrow3\left(x+2\right)-6\left(x-2\right)=-x\)
\(\Leftrightarrow3x+6-6x+12=-x\)
\(\Leftrightarrow3x-6x+x=-12-6\)
\(\Leftrightarrow-2x=-18\)
\(\Leftrightarrow x=9\left(nhận\right)\)
Vậy \(S=\left\{9\right\}\)
Câu 3 :
a, Xét ΔABD và ΔHBA có :
\(\widehat{A}=\widehat{H}=90^0\)
\(\widehat{B}:chung\)
\(\Rightarrow\Delta ABD\sim\Delta HBA\left(g-g\right)\)
b, Xét ΔADH và ΔDBC có :
\(\widehat{H}=\widehat{C}=90^0\)
\(\widehat{ADH}=\widehat{DBC}\left(AB//CD,slt\right)\)
\(\Rightarrow\Delta ADH\sim\Delta DBC\)
c, Ta có : \(\Delta ABD\sim\Delta HBA\left(cmt\right)\)
\(\Rightarrow\dfrac{AB}{BH}=\dfrac{BD}{AB}\)
\(\Rightarrow AB^2=BH.BD\)
d, Xét ΔABD vuông ở A , theo định lý Pi-ta-go ta được :
\(\Rightarrow BD=\sqrt{AB^2+AD^2}=\sqrt{12^2+9^2}=15\left(cm\right)\)
Ta có : \(\Delta ABD\sim\Delta HBA\left(cmt\right)\)
\(\Rightarrow\dfrac{AB}{BH}=\dfrac{BD}{AB}\)
hay \(\dfrac{12}{BH}=\dfrac{15}{12}\)
\(\Rightarrow BH=\dfrac{12.12}{15}=9,6\left(cm\right)\)