a) \(\Leftrightarrow x^2=\sqrt{4}\)

\(\Leftrightarrow x^2=2\Leftrightarrow x=\pm2\)

b) \(\Leftrightarrow\sqrt{\left(\dfrac{1}{2}x+1\right)^2}=9\)

\(\Leftrightarrow\left|\dfrac{1}{2}x+1\right|=9\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x+1=9\\\dfrac{1}{2}x+1=-9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=-16\end{matrix}\right.\)

c) \(\Leftrightarrow\sqrt{2x}-4\sqrt{2x}+16\sqrt{2x}=52\left(đk:x\ge0\right)\)

\(\Leftrightarrow13\sqrt{2x}=52\Leftrightarrow\sqrt{2x}=4\Leftrightarrow2x=16\Leftrightarrow x=8\left(tm\right)\)

f: Ta có: \(\sqrt{\dfrac{50-25x}{4}}-8\sqrt{2-x}+\sqrt{18-9x}=-10\)

\(\Leftrightarrow\sqrt{2-x}\cdot\dfrac{5}{2}-8\sqrt{2-x}+3\sqrt{2-x}=-10\)

\(\Leftrightarrow\sqrt{2-x}=4\)

\(\Leftrightarrow2-x=16\)

hay x=-14

Câu 31: 

#include <bits/stdc++.h>

using namespace std;

long long n,i,x,dem;

int main()

{

cin>>n;

dem=0;

for (i=1; i<=n; i++)

{

cin>>x;

if (30%x==0) dem++;

}

cout<<dem;

return 0;

}

\(a,\dfrac{6.12-6.7}{60}=\dfrac{6\left(12-7\right)}{60}=\dfrac{6.5}{60}=\dfrac{30}{60}=\dfrac{1}{2}\\ b,\dfrac{35.18-35}{\left(-34\right).7}=\dfrac{35\left(18-1\right)}{\left(-34\right).7}=\dfrac{35.17}{\left(-34\right).7}=\dfrac{595}{-238}=\dfrac{-5}{2}\)

\(c,\dfrac{42-11.42}{21.\left(-15\right)}=\dfrac{42\left(1-11\right)}{21.\left(-15\right)}=\dfrac{2.21.\left(-10\right)}{21.\left(-15\right)}=\dfrac{-20}{-15}=\dfrac{4}{3}\)

Câu 29:

a) nFe= 11,2/56=0,2(mol

nH2SO4=24,5/98=0,25(mol)

PTHH: Fe + H2SO4 -> FeSO4 + H2

Ta có: 0,2/1 < 0,25/1

=> nFe hết, nH2SO4 dư, tính theo nFe.

nH2SO4(p.ứ)=nH2=nFe=0,2(mol)

=>nH2SO4(dư)=nH2SO4(ban đầu) - nH2SO4(p.ứ)=0,25-0,2=0,05(mol)

=>mH2SO4(dư)=0,05.98=4,9(g)

b) V(H2,đktc)=0,2.22,4=4,48(l)

Câu 28:

-Cho quỳ tím nhận biết:+Quỳ tím chuyển đỏ:H2SO4

+Quỳ tím chuyển xanh:Ca(OH)2

+Quỳ tím không chuyển màu:NaNO3

Câu 29:

                 Fe+H2SO4--->FeSO4+H2

Theo PT:   1            1               1       1    mol

Theo đề bài:11,2    24,5                               g

Xét tỉ lệ:11,2/1             24,5/1=> cái nào nhiều hơn thì cái đó dư nhá.

b, Tính H2 theo  chất theo cái ko dư(Fe)   

1. adjust

2. design

3.

4. careful

5. memory

6. electronic

7. worried

8. famous

9. destroyed

10. selected

\(132-2x+8=46\Leftrightarrow140-2x=46\Leftrightarrow2x=94\Leftrightarrow x=47\)

\(2,\)

\(a,132-2\left(x-4\right)=46\)

\(\Rightarrow-2\left(x-4\right)=-86\)

\(\Rightarrow x-4=43\)

\(\Rightarrow x=47\)

Vậy: \(x=47\)

\(b.9x-4x=6^{17}:6^{15}+48:12\)

\(\Rightarrow5x=6^2+4\)

\(\Rightarrow5x=40\)

\(\Rightarrow x=8\)

Vậy: \(x=8\)

\(c,2^{x+3}.4=64\)

\(\Rightarrow2^{x+3}=16\)

\(\Rightarrow2^{x+3}=2^4\)

\(\Rightarrow x+3=4\)

\(\Rightarrow x=1\)

Vậy: \(x=1\)