{ x² - [ 6² - ( 8² - 9.7)³ - 7.5]³ - 5.3 }³ = 1

{ x² + [ 36 - (64 - 63)³ - 35]³ - 15}³ = 1

[ x² - ( 36 - 1³ - 35 ) - 15 ]³ = 1

[ x² - ( 36 - 1 - 35 ) - 15]³ = 1

[ x² - ( 35 - 35 ) - 15]³ = 1

[ x² - 0 - 15]³ = 1

( x² - 15 )³ = 1

<=> ( x² - 15)³ = 1³

=> x² - 15 = 1

<=> x² = 16

=> x = 4

\(\Leftrightarrow y\left(x+1\right)+2\left(x+1\right)+9=0\)

\(\Leftrightarrow\left(x+1\right)\left(y+2\right)=-9\)

Để x;y nguyên thì:

\(\left\{{}\begin{matrix}x+1=3\\y+2=-3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-5\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+1=-3\\y+2=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+1=1\\y+2=-9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-11\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+1=-9\\y+2=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=-1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+1=-1\\y+2=9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+1=9\\y+2=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=-3\end{matrix}\right.\)

\(x\left(y+2\right)+y+2=-9\Leftrightarrow\left(x+1\right)\left(y+2\right)=-9\)

\(\Rightarrow x+1;y+2\inƯ\left(-9\right)=\left\{\pm1;\pm3;\pm9\right\}\)

để\(\frac{2x-1}{3+x}\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}2x-1< 0\\3+x>0\end{cases}}\\\hept{\begin{cases}2x-1>0\\3+x< 0\end{cases}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}x< \frac{1}{2}\\x>-3\end{cases}\left(ktm\right)}\\\hept{\begin{cases}x>\frac{1}{2}\\x< -3\end{cases}\left(tm\right)}\end{cases}}\)

Vậy -3<x<1/2

Để 2x-1/3+x<0 

TH1 : 2x-1<0<=>2x<1<=>x<1/2

    và 3+x>0<=>x> -3 ( ktm)

TH2: 2x-1>0<=> 2x>1<=>x>1/2

    và 3+x<0<=>x< -3  (tm)

vậy -3<x<1/2

mk viết thường nên có chỗ nào ko hiểu thì ib cho mk nha nhớ đó

`#3107`

b)

`2.3^x = 162`

`\Rightarrow 3^x = 162 \div 2`

`\Rightarrow 3^x = 81`

`\Rightarrow 3^x = 3^4`

`\Rightarrow x = 4`

Vậy, `x = 4`

c)

`(2x - 15)^5 = (2 - 15)^3`

\(\Rightarrow \)`(2x - 15)^5 - (2x - 15)^3 = 0`

\(\Rightarrow \)`(2x - 15)^3 . [ (2x - 15)^2 - 1] = 0`

\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=15\\\left(2x-15\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x-15=1\\2x-15=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x=16\\2x=-14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=-7\end{matrix}\right.\)

Vậy, `x \in`\(\left\{-7;8;\dfrac{15}{2}\right\}.\)

`d)`

\(3^{x+2}-5.3^x=?\) Bạn ghi tiếp đề nhé!

`e)`

\(7\cdot4^{x-1}+4^{x-1}=23?\)

\(4^{x-1}\cdot\left(7+1\right)=23\\ \Rightarrow4^{x-1}\cdot8=23\\ \Rightarrow4^{x-1}=\dfrac{23}{8}\)

Bạn xem lại đề!

`f)`

\(2\cdot2^{2x}+4^3\cdot4^x=1056\)

\(\Rightarrow2\cdot2^{2x}+\left(2^2\right)^3\cdot\left(2^2\right)^x=1056\\ \Rightarrow2\cdot2^{2x}+2^6\cdot2^{2x}=1056\\ \Rightarrow2^{2x}\cdot\left(2+2^6\right)=1056\\ \Rightarrow2^{2x}\cdot66=1056\\ \Rightarrow2^{2x}=1056\div66\\ \Rightarrow2^{2x}=16\\ \Rightarrow2^{2x}=2^4\\ \Rightarrow2x=4\\ \Rightarrow x=2\)

Vậy, `x = 2`

_____

\(10 -{[(x \div 3+17) \div 10+3.2^4] \div 10}=5\)

\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)

\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)

\(\Rightarrow\left(x\div3+17\right)\div10+48=50\)

\(\Rightarrow\left(x\div3+17\right)\div10=2\)

\(\Rightarrow x\div3+17=20\)

\(\Rightarrow x\div3=3\\ \Rightarrow x=9\)

Vậy, `x = 9.`

Ta có

\(\sqrt{-x^2+2x+2}=\sqrt{-x^2+2x-1+3}=\sqrt{-\left(x-1\right)^2+3}\le\sqrt{3}\)

\(\sqrt{-x^2-6x-8}=\sqrt{-x^2-6x-9+1}=\sqrt{-\left(x+3\right)^2+1}\le1\)

\(\Rightarrow\sqrt{-x^2+2x+2}+\sqrt{-x^2-6x-8}\le1+\sqrt{3}\)

Dấu "=" xảy ra khi x-1=0 và x+3=0 nên x=1  và x=-3(VL). Phương trình vô nghiệm

Ta có :   D=(x^2 -  2x  +  1)  +  (4x^2 +  4x +1)

                 = x^2 - 2x + 1 + 4x^2 +4x +  1

                 = 5x^2 + 2x + 2

                 =5(x^2  +  2/5x + 2/5)

                 =5(x^2 + 2/5x + 1/25 + 9/25)

                 =5(x^2 +2/5x +1/25) + 9/5 >= 9/5

 Vậy MinD=9/5 khi x=-1/5

Mình bổ sung 1 chút ở chỗ 5(x^2 + 2/5x + 1/25) + 9/5 = 5(x+1/5)^2  + 9/5 >= 9/5

Chỗ cuối bạn có thể giải thích rõ hơn đc k hả bạn

  nếu mà x<0 

bạn bấm thiếu đề rồi kìa

sai chứ không phải thiếu mình lộn

6143/15345

=>x-9>0 hoặc x-3<0

=>x<3 hoặc x>9

a) 

5.(12-x)-20=30

⇒60-5x-20=30

⇒-5x=30+20-60

⇒-5x=-10

⇒x=2

b)(17x - 25 ) : 8 + 65 = 9²

(17x - 25 ) : 8 + 65 = 81

17x - 25 = 16 x 8 = 128

17x = 128+25=153

x= 153:17 =9

c)

x=23

Giải thích các bước giải:

3x – 10 = 2x + 13

3x-2x=13+10

x=23

d)4(2x+7)-3(3x-2)=24

4.2x+4.7-3.3x+3.2=24

8x+28-9x+6=24

8x-9x=24-28-6=-10

=>(-1)x=-10

        x=-10:(-1)

        x=10

a. \(5\cdot\left(12-x\right)-20=30\Leftrightarrow5\left(12-x\right)=50\)

\(\Leftrightarrow12-x=50:5=10\)

\(\Leftrightarrow x=12-10=2\)

b. \(\left(17x-25\right):8+65=9^2\)

\(\Leftrightarrow\left(17x-25\right):8=81-65=16\)

\(\Leftrightarrow17x-25=16:8=2\)

\(\Leftrightarrow17x=2+25=27\Leftrightarrow x=\frac{27}{17}\)

c. \(3x-10=2x+13\)

\(\Leftrightarrow3x-2x=10+13\)

\(\Leftrightarrow x=23\)

d. \(4\cdot\left(2x+7\right)-3\cdot\left(3x-2\right)=24\)

\(\Leftrightarrow8x+28-9x+6=24\)

\(\Leftrightarrow34-x=24\Leftrightarrow x=10\)

a) 

5.(12-x)-20=30

⇒60-5x-20=30

⇒-5x=30+20-60

⇒-5x=-10

⇒x=2

b)(17x - 25 ) : 8 + 65 = 9²

(17x - 25 ) : 8 + 65 = 81

17x - 25 = 16 x 8 = 128

17x = 128+25=153

x= 153:17 =9

c)

x=23

Giải thích các bước giải:

3x – 10 = 2x + 13

3x-2x=13+10

x=23

d)4(2x+7)-3(3x-2)=24

4.2x+4.7-3.3x+3.2=24

8x+28-9x+6=24

8x-9x=24-28-6=-10

=>(-1)x=-10

        x=-10:(-1)

        x=10