\(3^{x-1}.7+3^{x-1}.2=9\\ 3^{x-1}.\left(7+2\right)=9\\ 3^{x-1}.9=9\\ 3^{x-1}=\dfrac{9}{9}=1\\ Mà:3^0=1\\ Nên:x-1=0\\ Vậy:x=0+1=1\\ ---\\ P=2+2^2+2^3+...+2^{65}+2^{66}=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{64}+2^{65}+2^{66}\right)\\ =2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{64}\left(1+2+2^2\right)\\ =2.7+2^4.7+...+2^{64}.7\\ =\left(2+2^4+....+2^{64}\right).7⋮7\left(đpcm\right)\)

+)
\(3^{x-1}.7+3^{x-1}.2=9\)
\(3^{x-1}.\left(7+2\right)=9\)
\(3^{x-1}.9=9\)
\(3^{x-1}=9:9\)
\(3^{x-1}=1\)
⇔\(3^{x-1}=3^0\)
⇒\(x-1=0\)
\(x=0+1\)
\(x=1\)
Vậy \(x=1\)

+)
\(2+2^2+2^3+...+2^{65}+2^{66}\)
Vì \(2+2^2+2^3=14\) mà \(14\)⋮\(7\)
⇒Ta nhóm 3 số với nhau
Ta có:
\(\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{64}+2^{65}+2^{66}\right)\)
\(\left(2+2^2+2^3\right)+2^3.\left(2+2^2+2^3\right)+...+2^{63}.\left(2+2^2+2^3\right)\)
\(14.1+14.2^3+...+14.2^{63}\)
\(14.\left(1+2^3+...+2^{63}\right)\)
Do \(14\)⋮\(7\) nên \(P=14.\left(2+2^3+...+2^{63}\right)\)⋮\(7\)

Xin tick

\(\left(3x-7\right)^3=2^3.3^2+53\)

\(\left(3x-7\right)^3=8.9+53=125\)

\(\Leftrightarrow\left(3x-7\right)^3=125=5^3\)

\(\Leftrightarrow3x-7=5\)

\(\Leftrightarrow3x=5+7\)

\(\Leftrightarrow3x=12\)

\(\Leftrightarrow x=12:3\)

\(\Leftrightarrow x=4\)

${(3x-7)}^3=2^3.3^2+53$

$\iff {(3x-7)}^3=8.9+53$

$\iff {(3x-7)}^3=72+53$

$\iff {(3x-7)}^3=125$

$\iff 3x-7=5$

$\iff 3x=5+7$

$\iff 3x=12$

$\iff x=12:3$

$\iff x=4$

Vậy $x=4$

$\mathrm{đây\; em\; nhé}$

\(\Leftrightarrow\left(3x-7\right)^3=8.9+53=125=5^3\)

\(\Rightarrow3x-7=5\Leftrightarrow3x=12\Leftrightarrow x=4\)

b) ( 3x – 2 3 ) . 7 = 7 4

3x – 8 =  7 4 : 7

3x – 8 = 7 3

3x – 8 = 343

3x = 343 + 8

3x = 351

x = 351 : 3 = 117

a) (x – 45).27 = 0

=>   x -  45  =  0

=>  x  =  45

 b) 23.(42- x) = 23 

=>  42- x  =  1

=>   x  =  41

c. 3x – 5=7

=>  3x  =  12

=>  x  =  4

e. 15 – 5x=10

=>  5x  =  5

=>  x  =  1

a) \(A=\sqrt{3x-5}+\sqrt{7-3x}\)

Áp dụng bất đẳng thức Bunhiacopxki, ta có ; \(A^2=\left(1.\sqrt{3x-5}+1.\sqrt{7-3x}\right)^2\le\left(1^2+1^2\right)\left(3x-5+7-3x\right)\)

\(\Rightarrow A^2\le4\Rightarrow A\le2\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{5}{3}\le x\le\frac{7}{3}\\\sqrt{3x-5}=\sqrt{7-3x}\end{cases}\Leftrightarrow x=2}\)

Vậy Max A = 2 <=> x= 2

b) \(B=\sqrt{x-5}+\sqrt{23-x}\)

Áp dụng bất đẳng thức Bunhiacopxki, ta có ; \(B^2=\left(1.\sqrt{x-5}+1.\sqrt{23-x}\right)^2\le\left(1^2+1^2\right)\left(x-5+23-x\right)\)

\(\Rightarrow B^2\le36\Leftrightarrow B\le6\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}5\le x\le23\\\sqrt{x-5}=\sqrt{23-x}\end{cases}\Leftrightarrow}x=14\)

Vậy Max B = 6 <=> x = 14

cho mình xin cái link đc k, sao mình không thấy được câu trả lời của bạn

LINK đây bạn nhé ^^

http://olm.vn/hoi-dap/question/631449.html

\(\left(3x+2\right)\left(x-1\right)+\left(x+3\right)\left(x-7\right)+2x+23=0\\ \Leftrightarrow3x^2+2x-3x-2+x^2+3x-7x-21+2x+23=0\\ \Leftrightarrow3x^2-x^2+2x-3x+3x-7x+2x-2-21+23=0\\ \Leftrightarrow x^2-3x=0\\ \Leftrightarrow x.\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)

<=> x=0 hoặc x=3

(3x+2)(x-1)+(x+3)(x-7)+2x+23=0

=>3x²+2x-3x-2+x²+3x-7x-21+2x=-23

=>(3x²+x²)+(2x-3x+3x-7x+2x) -(2+21)=-23

=>4x²-3x-23=-23

=>4x²-3x=-23+23=0

=>x(4x-3)=0

=>x=0 hoặc 4x-3=0

=>x=0 hoặc x=3/4.

\(\Leftrightarrow\frac{11}{35}

Bài 22:

a: =>-12x+60+21-7x=5

=>-19x+81=5

=>-19x=-76

=>x=4

b: =>30x+60-6x+30-24x=100

=>90=100(loại)

-27 + 5( x-7 ) = 23 - (71-3x)

-27 + 5( x-7 ) = 23 - 71 + 3x

5(x-7) - 3x = 23 - 71 + 27

5x - 35 - 3x = -21

x( 5-3 ) - 35 = -21

x.2 - 35 = -21

x.2 = -21 + 35

x.2 = 14

x = 7

Vậy x = 7

-27+5(x-7)=23-(71-3x)
=>-27+5x-35=23-71+3x
=>-27-35-23+71=3x-5x
=>-14=-2x
=>-2x=-14
=>x=-14:(-2)
=>x=7
Vậy x=7

Dễ vậy ,tự làm đi bn