viet cach lam luon nha

Ta có:\(2^{36}\)và \(3^{27}\)

\(2^{36}=\left(2^4\right)^9=16^9\)

\(3^{27}=\left(3^3\right)^9=27^9\)

Vì \(16< 27\Rightarrow16^9< 27^9\)

Vậy....

b,\(9^{20}\)và \(9999^{10}\)

\(9^{20}=\left(9^2\right)^{10}=81^{10}\)

\(9999^{10}\)

Vì \(81< 9999\Rightarrow81^{10}< 9999^{10}\)

Vậy ...

c,\(54^4\)

\(21^{12}=\left(21^3\right)^4=9261^4\)

Vì \(54< 9261\Rightarrow54^4< 9261^4\)

Vậy...

cam on

Không đáp án nào đúng

ko đáp án nào đúng cả

7x-18+9x=4x+54+24

7x+9x=4x+78+18

7x+9x=4x+96

16x-4x=96

12x=96

x=96:12

x=8

7x-18+9x=4x+78

7x+9x=4x+78+18

7x+9x=4x+96

16x-4x=96

12x=96

x=96:12

x=8

Bài 1:

a) \(8^5\cdot8^2=8^7\)

b) \(9^3\cdot3^2=\left(3^2\right)^3\cdot3^2=3^6\cdot3^2=3^8\)

c) \(2^7\cdot5^7=10^7\)

d) \(27^6:3^3=\left(3^3\right)^6:3^3=3^{18}:3^3=3^{15}\)

Bài 2:

a) \(x^6:x^3=125\)

\(\Rightarrow x^3=125\)

\(\Rightarrow x=5\)

b) \(x^{20}=x\)

\(\Rightarrow x^{20}-x=0\)

\(\Rightarrow x\left(x^{19}-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{19}-1=0\Rightarrow x=1\end{matrix}\right.\)

c) \(3^x\cdot3=243\)

\(\Rightarrow3^x=81\)

\(\Rightarrow x=4\)

d) \(2x-138=2^3\cdot3^2\)

\(\Rightarrow2x-138=72\)

\(\Rightarrow2x=200\)

\(\Rightarrow x=100\)

Giải:

Bài 1:

a) \(8^5.8^2=8^{5+2}=8^7\)

b) \(9^3.3^2=3^6.3^2=3^{6+2}=3^8\)

c) \(2^7.5^7=\left(2.5\right)^7=10^7\)

d) \(27^6:3^3=3^{18}:3^3=3^{18-3}=3^{15}\)

Bài 2:

a) \(x^6:x^3=x^{6-3}=x^3=125\)

\(\Leftrightarrow x=5\)

b) \(x^{20}=x\)

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

c) \(3^x.3=243\)

\(\Leftrightarrow3^{x+1}=243\)

\(\Leftrightarrow3^{x+1}=3^5\)

\(\Leftrightarrow x+1=5\Leftrightarrow x=4\)

d) \(2.x-138=2^3.3^2\)

\(\Leftrightarrow2.x-138=8.9\)

\(\Leftrightarrow2.x-138=72\)

\(\Leftrightarrow2.x=72+138\)

\(\Leftrightarrow2.x=210\Leftrightarrow x=105\)

Chúc bạn học tốt!

Bài 1 : Viết kết quả sau dưới dạng một lũy thừa :

a) 8⁵ . 8² = 8⁷

b) 9³ . 3² = 9³ . 9 = 9⁴

c) 2⁷ . 5⁷ = 10⁷

d) 27⁶ : 3³ = 27⁶ : 27 = 27⁵

Bài 2 : Tìm x biết :

a) x⁶ : x³ = 125

\(\Rightarrow\) x³ = 5³

\(\Rightarrow\) x = 5.

b) x²⁰ = x

- Nếu x = 0 thì 0²⁰ = 0 (chọn)

- Nếu x = 1 thì 1²⁰ = 1 (chọn)

- Nếu x \(\ge\) 2 thì x²⁰ = x (vô lí)

Vậy x \(\in\) {0; 1}.

c) 3^x . 3 = 243

\(\Rightarrow\) 3^x . 3 = 3⁵

3^x = 3⁵ : 3

3^x = 3⁴

\(\Rightarrow\) x = 4.

d) 2x - 138 = 2³ . 3²

2x - 138 = 8 . 9

2x - 138 = 72

2x = 72 + 138

2x = 210

x = 210 : 2

x = 105.

Bài 3 : So sánh :

c) 4⁴ < 64⁷

Bài 4 : Cho n! =1.2.3.4...........n . Tính 5! - 4!.

5! - 4! = 120 - 24 = 96.

\(\frac{1\cdot3\cdot9+2\cdot6\cdot18+3\cdot9\cdot27}{1\cdot5\cdot18+2\cdot10\cdot36+3\cdot15\cdot54}\)

\(=\frac{1\cdot3\cdot9+2\left(1\cdot3\cdot9\right)+3\left(1\cdot3\cdot9\right)}{1\cdot5\cdot18+2\left(1\cdot5\cdot18\right)+3\left(1\cdot5\cdot18\right)}\)

\(=\frac{\left(1\cdot3\cdot9\right)\left(1+2+3\right)}{\left(1\cdot5\cdot18\right)\left(1+2+3\right)}\)

\(=\frac{3}{10}\)

a: \(\Leftrightarrow x^2\left(9x^2-4\right)=0\)

\(\Leftrightarrow x^2\left(3x-2\right)\left(3x+2\right)=0\)

hay \(x\in\left\{0;\dfrac{2}{3};-\dfrac{2}{3}\right\}\)

b: \(\Leftrightarrow2x^4-4x^2+3x^2-6=0\)

\(\Leftrightarrow x^2-2=0\)

hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)

d: \(\Leftrightarrow x^4-9x^2+6x^2-54=0\)

\(\Leftrightarrow x^2-9=0\)

=>x=3 hoặc x=-3

b, \(đk:x\ge2\)

Xét x=2 thay vào pt thấy không thỏa mãn => x>2 hay 27x-54>0

 \(x^3-11x+36x-18=4\sqrt[4]{27x-54}\)

\(\Leftrightarrow27x^3-297x^2+972x-486=4\sqrt[4]{\left(27x-54\right).81.81.81}\le189+27x\) (cosi với 4 số dương, dấu = xảy ra khi x=5)

\(\Leftrightarrow x^3-11x^2+35x-25\le0\)

\(\Leftrightarrow\left(x-1\right)\left(x-5\right)^2\le0\)  (*)

Có \(\left\{{}\begin{matrix}x>2\\\left(x-5\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1>0\\\left(x-5\right)^2\ge0\end{matrix}\right.\)\(\Rightarrow\left(x-1\right)\left(x-5\right)^2\ge0\) (2*)

Từ (*) và (2*) ,dấu = xra khi x=5 (thỏa mãn)
Vây pt có nghiệm duy nhất x=5

c,Có \(6\sqrt[3]{4x^3+x}=16x^4+5>0\)

\(\Leftrightarrow4x^3+x>0\)

Có: \(16x^4+5=6\sqrt[3]{4x^3+x}\le2\left(4x^3+x+2\right)\) (theo cosi với 3 số dương,dấu = xảy ra khi \(x=\dfrac{1}{2}\))

\(\Leftrightarrow16x^4-8x^3-2x+1\le0\)

\(\Leftrightarrow\left(2x-1\right)^2\left(4x^2+2x+1\right)\le0\) (*)
(tương tự câu b) Dấu = xảy ra khi \(x=\dfrac{1}{2}\)(thỏa mãn)
Vậy....

d) Đk: \(x\ge\dfrac{3}{4}\)

Áp dụng bđt cosi:

 \(\sqrt{2x-1}\le\dfrac{2x-1+1}{2}=x\)

 \(\Rightarrow\dfrac{1}{\sqrt{2x-1}}\ge\dfrac{1}{x}\) (*)

\(\sqrt[4]{4x-3}\le\dfrac{4x-3+1+1+1}{4}=x\)

\(\dfrac{\Rightarrow1}{\sqrt[4]{4x-3}}\ge\dfrac{1}{x}\) (2*)

Từ (*) và (2*) \(\Rightarrow\dfrac{1}{\sqrt{2x-1}}+\dfrac{1}{\sqrt[4]{4x-3}}\ge\dfrac{2}{x}\)

Dấu = xảy ra khi x=1 (tm)

`a)\sqrtx+\sqrt{2-x}=(3x^2-2x+3)/(x^2+1)`

`đk:0<=x<=2`

`pt<=>sqrtx-1+\sqrt{2-x}-1=(3x^2-2x+3)/(x^2+1)-2`

`<=>(x-1)/(sqrtx+1)+(1-x)/(sqrt{2-x}+1)=(x^2-2x+1)/(x^2+1)`

`<=>(x-1)/(sqrtx+1)+(1-x)/(sqrt{2-x}+1)=(x-1)^2/(x^2+1)`

`<=>(x-1)((x-1)/(x^2+1)+1/(sqrt{2-x}+1)-1/(sqrtx+1))=0`

`<=>x-1=0<=>x=1`

Vậy `S={1}`