2^3+x=32

2^3+x=2^4

Ta có:

3+x=4

X=4-3

X=1 

`#3107.101107`

c)

`2^(x + 3) = 32`

`=> 2^(x + 3) = 2^5`

`=> x + 3 = 5`

`=> x = 5 - 3`

`=> x = 2`

Vậy, `x = 2`

d)

`20 - 2(x - 1)^2 = 2`

`=> 2(x - 1)^2 = 20 - 2`

`=> 2(x - 1)^2 = 18`

`=> (x - 1)^2 = 18 \div 2`

`=> (x - 1)^2 = 9`

`=> (x - 1)^2 = (+-3)^2`

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

`=>`\(\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

Vậy, `x \in {-2; 4}.`

____

b)

`(2x) \in` Ư`(20)` và `x > 8`

Ta có:

Ư`(20) = {1; 2; 3; 4; 5; 10; 20}`

`=> 2x \in {1; 2; 3; 4; 5; 10; 20}`

`=> x \in {1/2; 1; 3/2; 2; 5/2; 5; 10}`

Mà `x > 8`

`=> x = 10`

Vậy, `x = 10`

c)

`30 \vdots (x + 2)` và `x \le 15`

Ta có:

Ư`(30) = {1; 2; 3; 5; 6; 10; 15; 30}`

`=> (x + 2) \in {1; 2; 3; 5; 6; 10; 15; 30}`

`=> x \in {-1; 0; 1; 3; 4; 8; 13; 28}`

Mà `x \le 15 => x \in {-1; 0; 1; 3; 4; 8}`

Vậy, `x \in {-1; 0; 1; 3; 4; 8}.`

a) \(7^2-7\left(13-x\right)=14\)

\(7\left(13-x\right)=49-14=35\)

\(13-x=5\)

\(x=13-5=8\)

b) \(5x-5^2=10\)

\(5x=10+25=35\)

\(x=7\)

c) \(4\left(x-5\right)-2^3=2^4.3=48\)

\(4\left(x-5\right)=48+8=56\)

\(x-5=14\)

\(x=19\)

4:

a: \(\Leftrightarrow49+7\left(x-13\right)=14\)

=>7(x-13)=35

=>x-13=5

=>x=18

b: \(5x-5^2=10\)

=>\(5x=10+25=35\)

=>x=7

c: \(4\left(x-5\right)-2^3=2^4\cdot3\)

=>\(4\left(x-5\right)=16\cdot3+8=56\)

=>x-5=14

=>x=19

Câu 4 : 

\(n_{Fe2O3}=\dfrac{24}{160}=0,15\left(mol\right)\)

Pt : \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O|\)

          1               3                    1                 3

       0,15                                 0,15

a) \(n_{Fe2\left(SO4\right)3}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)

⇒ \(m_{Fe2\left(SO4\right)3}=0,15.400=60\left(g\right)\)

b) \(C_{M_{Fe2\left(SO4\right)3}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\)

 Chúc bạn học tốt

a,\(n_{Fe_2O_3}=\dfrac{24}{160}=0,15\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂SO₄ → Fe₂(SO₄)₃ + 3H₂O

Mol:     0,15         0,45                0,15                   

\(m_{Fe_2\left(SO_4\right)_3}=0,15.400=60\left(g\right)\)

b,\(C_{M_{ddFe_2\left(SO_4\right)_3}}=\dfrac{0,15}{0,5}=0,3\left(mol\right)\)

Fe2O3 + 3H2SO4 → Fe2(SO4)3 + 3H2O

0,15--------------------------------0,15 mol

n Fe2O3=24\160=0,15 mol

=>m Fe2(SO4)3=0,15.400=60g

mà thể tích khôi đổi 

=>CM Fe2(SO4)3=0,15\0,5=0,3 M

Câu 9:

a) Ta có: \(9x^2-16=0\)

\(\Leftrightarrow\left(3x-4\right)\left(3x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)

b) Ta có: \(4x^2=13\)

\(\Leftrightarrow x^2=\dfrac{13}{4}\)
\(\Leftrightarrow x\in\left\{\dfrac{\sqrt{13}}{2};-\dfrac{\sqrt{13}}{2}\right\}\)

c) Ta có: \(2x^2+9=0\)

\(\Leftrightarrow2x^2=-9\)(Vô lý)

d) Ta có: \(-x^2+324=0\)

\(\Leftrightarrow x^2=324\)

\(\Leftrightarrow\left[{}\begin{matrix}x=18\\x=-18\end{matrix}\right.\)

1C 2B 3D 4B 5A 

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