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7 tháng 10 2021

1 A

2 B

3 A

4 C

D

1 C

2 D

3 A

4 D

5 B

7 D

8 B

9 C

10 C

E

1 C

2 C

3 A

4 B

5 D

5 tháng 10 2021

Câu 53: C
Câu 55 C

Câu 56 C

Câu 59C

30 tháng 9 2021

Bài 1:

\(n_{CuO}=\dfrac{56}{80}=0,7\left(mol\right)\)

PTHH: CuO + 2HCl → CuCl2 + H2O

Mol:      0,7       1,4

\(m_{ddHCl}=\dfrac{1,4.36,5.100}{14,6}=350\left(g\right)\)

Bài 2:

\(n_{Na_2SO_3}=\dfrac{12,6}{126}=0,1\left(mol\right)\)

PTHH: Na2SO3 + 2HCl → 2NaCl + SO2 + H2O

Mol:         0,1                                      0,1

\(V_{SO_2}=0,1.22,4=2,24\left(l\right)\)

25 tháng 9 2021

1) \(\sqrt{2x-5}=7\)

\(\left(\sqrt{2x-5}\right)^2=7^2\)

\(2x-5=49\)

\(2x=54\)

\(x=27\)

2) \(3+\sqrt{x-2}=4\)

\(\sqrt{x-2}=1\)

\(\left(\sqrt{x-2}\right)^2=1^2\)

\(x-2=1\)

\(x=3\)

25 tháng 9 2021

1) \(\sqrt{2x-5}=7\left(đk:x\ge\dfrac{5}{2}\right)\)

\(\Leftrightarrow2x-5=49\Leftrightarrow2x=54\Leftrightarrow x=27\left(tm\right)\)

2) \(3+\sqrt{x-2}=4\left(đk:x\ge2\right)\)

\(\Leftrightarrow\sqrt{x-2}=1\Leftrightarrow x-2=1\Leftrightarrow x=3\)

3) \(\Leftrightarrow\sqrt{\left(x-1\right)^2}=1\Leftrightarrow\left|x-1\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

4) \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=1\Leftrightarrow\left|x-2\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

5) \(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x+4\right)^2}\)

\(\Leftrightarrow\left|2x-1\right|=\left|x+4\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+4\\2x-1=-x-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

6) \(ĐK:x\ge-2\)

 \(\Leftrightarrow5\sqrt{x+2}-3\sqrt{x+2}-\sqrt{x+2}=\sqrt{x+7}\)

\(\Leftrightarrow\sqrt{x+2}=\sqrt{x+7}\)

\(\Leftrightarrow x+2=x+7\Leftrightarrow2=7\left(VLý\right)\)

Vậy \(S=\varnothing\)

7) \(ĐK:x\ge-1\)

\(\Leftrightarrow5\sqrt{2x+1}+3\sqrt{x+1}=4\sqrt{x+1}+4\sqrt{2x+1}\)

\(\Leftrightarrow\sqrt{2x+1}=\sqrt{x+1}\)

\(\Leftrightarrow2x+1=x+1\Leftrightarrow x=0\left(tm\right)\)

8 tháng 12 2023

câu a, \(\dfrac{x}{x+1}\)\(\dfrac{x^2}{1-x}\)\(\dfrac{1}{x^2-1}\)  (đk \(x\)≠ -1; 1)

          \(x^2\) - 1 = ( \(x\) - 1).(\(x\) + 1)

          \(\dfrac{x}{x+1}\) = \(\dfrac{x.\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}\);

          \(\dfrac{x^2}{1-x}\) = \(\dfrac{-x^2}{x-1}\)\(\dfrac{-x^2.\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\) 

         \(\dfrac{1}{x^2-1}\)  =  \(\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)

b, \(\dfrac{10}{x+2}\)\(\dfrac{5}{2x-4}\)\(\dfrac{1}{6-3x}\) (đk \(x\) ≠ -2; 2)

    2\(x-4\) = 2.(\(x\) - 2); 6 - 3\(x\) = - 3.(\(x\)  - 2)

   \(\dfrac{10}{x+2}\) = \(\dfrac{10.2.3\left(x-2\right)}{2.3\left(x+2\right)\left(x-2\right)}\) = \(\dfrac{60\left(x-2\right)}{6\left(x-2\right)\left(x+2\right)}\)

    \(\dfrac{5}{2x-4}\) = \(\dfrac{5.3\left(x+2\right)}{2.3\left(x-2\right).\left(x+2\right)}\) = \(\dfrac{15.\left(x+2\right)}{6.\left(x-2\right)\left(x+2\right)}\)

    \(\dfrac{1}{6-3x}\) = \(\dfrac{-1}{3.\left(x-2\right)}\) = \(\dfrac{-1.\left(x+2\right)}{3.2.\left(x-2\right)\left(x+2\right)}\) = \(\dfrac{-2.\left(x+2\right)}{6.\left(x-2\right).\left(x+2\right)}\)

   

         

 

8 tháng 12 2023

c, \(\dfrac{x}{2x-4}\)\(\dfrac{1}{2x+4}\) và \(\dfrac{3}{4-x^2}\)  đk \(x\) ≠ 2; -2

\(\dfrac{x}{2x-4}\)  =   \(\dfrac{x}{2.\left(x-2\right)}\) = \(\dfrac{x.\left(x+2\right)}{2.\left(x-2\right).\left(x+2\right)}\) 

  \(\dfrac{1}{2x+4}\) = \(\dfrac{1}{2.\left(x+2\right)}\) = \(\dfrac{\left(x-2\right)}{2.\left(x+2\right).\left(x-2\right)}\)

\(\dfrac{3}{4-x^2}\) = \(\dfrac{-3}{\left(x-2\right)\left(x+2\right)}\)  = \(\dfrac{-6}{2.\left(x-2\right)\left(x+2\right)}\)

 

9:

a: XétΔABC vuông tại A và ΔHBA vuông tại H có

góc B chung

=>ΔABC đồng dạng với ΔHBA

=>BA/BH=BC/BA
=>BA^2=BH*BC

b: BC=25cm; AB=căn 9*25=15cm; AC=căn 16*25=20cm

S ABC=1/2*15*20=150cm2

C ABC=25+15+20=60cm

12 tháng 5 2022

he would join the Science Club the day after
his room is untidy, his mother is unhappy
started our work, she had explained everything clearly
I had enough money, I could buy this motorbike 

11 tháng 12 2023

Nhìn ngợp z

AH
Akai Haruma
Giáo viên
5 tháng 1

Lời giải:

ĐKXĐ: $x\geq 0; x\neq 9$

\(C=\frac{\sqrt{x}(\sqrt{x}+3)-(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{x+3}\\ =\frac{x+2\sqrt{x}+3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{x+3}=\frac{x+2\sqrt{x}+3}{(\sqrt{x}+3)(x+3)}\)