a) 232 - (3x + 2).5 = 2³.5²

232 - (3x + 2).5 = 8.25

232 - 5.(3x + 2) = 200

-5.(3x + 2) = 200 - 232

-5.(3x + 2) = -32

3x + 2 = (-32) : (-5)

3x + 2 = 32/5

3x = 32/5 - 2

3x = 22/5

x = 22/5 : 3

x = 22/15

\(a)232-\left(3x+2\right).5=2^3.5^2\)

\(\Leftrightarrow232-\left(3x+2\right).5=200\)

\(\Leftrightarrow\left(3x+2\right).5=32\)

\(\Leftrightarrow3x+2=6.4\)

\(\Leftrightarrow3x=4.4\)

\(\Leftrightarrow\frac{22}{15}\)

\(b)3.5^x-10^0=74\)

\(\Leftrightarrow3.5^x=75\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow5^x=5^2\)

\(\Leftrightarrow x=2\)

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

\(\Leftrightarrow4^{2x+4}=4^3\)

\(\Leftrightarrow2x+4=3\)

\(\Leftrightarrow x=-\frac{1}{2}\)

\(d)5^{x-1}+5^x=100\)

\(\Leftrightarrow5^x.\frac{1}{5}+5^x=100\)

\(\Leftrightarrow5^x\left(\frac{1}{5}+1\right)=100\)

\(\Leftrightarrow5^x=\frac{250}{3}\)

Chưa nghĩ ra. Các câu khác tương tự như các câu trên.

`@` `\text {Ans}`

`\downarrow`

Gửi c!

Bài 1: 

a) \(3x^2\left(2x^3-x+5\right)-6x^5-3x^3+10x^2\)

\(=6x^5-3x^3+10x^2-6x^5-3x^3+10x^2\)

\(=10x^2+10x^2\)

\(=20x^2\)

b) \(-2x\left(x^3-3x^2-x+11\right)-2x^4+3x^3+2x^2-22x\)

\(=-2x^4+6x^3+2x^2-22x-2x^4+3x^3+2x^2-22x\)

\(=-4x^4+9x^3+4x^2-44x\)

Câu hỏi đâu ạ ? 

câu hỏi điu

\(\left(x-2\right)\left(x^2+2x+4\right)+3x-4=\left(x+2\right)\left(x^2-2x+4\right)-x+1\)

\(\Rightarrow\left(x^3-8\right)+3x-4=\left(x^3+8\right)-x+1\)

\(\Rightarrow x^3-8+3x-4=x^3+8-x+1\)

\(\Rightarrow x^3-x^3+3x+x=8+8+4+1\)

\(\Rightarrow4x=21\)

\(\Rightarrow x=\dfrac{21}{5}\)

Nhân 3x chứ đâu phải cộng 3x đâu c.

Bài 2: 

Sửa đề:  \(x^3-3x^2-10x=0\)

\(\Leftrightarrow x\left(x^2-3x-10\right)=0\)

\(\Leftrightarrow x\left(x-5\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-2\end{matrix}\right.\)

a) (x-2)³+6(x+1)²-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6x²+12x+6-x³+12=0

\(\Rightarrow\)24x+10=0

\(\Rightarrow\)24x=-10

\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)

b)(x-5)(x+5)-(x+3)²+3(x-2)²=(x+1)²-(x-4)(x+4)+3x²

\(\Rightarrow\)x²-25-(x²+6x+9)+3(x²-4x+4)=x²+2x+1-(x²-16)+3x²

\(\Rightarrow\)x²​-25-x²-6x-9+3x²-12x+12=x²+2x+1-x²+16+3x²

\(\Rightarrow\)3x²-18x-22=3x²+2x+17

\(\Rightarrow\)3x²-18x-22-3x²-2x-17=0

\(\Rightarrow\)-20x-39=0

\(\Rightarrow\)-20x=39

\(\Rightarrow\)x=\(-\dfrac{39}{20}\)

1) ĐKXĐ: \(x\notin\left\{-2;2\right\}\)

Ta có: \(\dfrac{x-1}{x+2}-\dfrac{9}{x^2-4}=\dfrac{-3}{x-2}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{9}{\left(x-2\right)\left(x+2\right)}=\dfrac{-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2-3x+2-9=-3x-6\)

\(\Leftrightarrow x^2-3x-7+3x+6=0\)

\(\Leftrightarrow x^2-1=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\)

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

Vậy: S={1;-1}

2)

Sửa đề: \(\dfrac{3x-3}{x^2-9}-\dfrac{1}{x-3}=\dfrac{x+1}{x+3}\)

ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

Ta có: \(\dfrac{3x-3}{x^2-9}-\dfrac{1}{x-3}=\dfrac{x+1}{x+3}\)

\(\Leftrightarrow\dfrac{3x-3}{\left(x-3\right)\left(x+3\right)}-\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)

Suy ra: \(3x-3-x-3=x^2-3x+x-3\)

\(\Leftrightarrow x^2-2x-3=2x-6\)

\(\Leftrightarrow x^2-2x-3-2x+6=0\)

\(\Leftrightarrow x^2-4x+3=0\)

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

\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)

Vậy: S={1}

`1)(x-1)/(x+2)-9/(x^2-4)=-3/(x-2)(x ne 2)`

`<=>x^2-3x+2-9=-3x-6`

`<=>x^2-1=0`

`<=>x=+-1`