a: =>14x+20+5=6x-9-9x

=>14x+25=-3x-9

=>17x=-34

=>x=-2

b: =>\(2x^2-30x+2x-30=2x^2+10x-10x-50\)

=>-28x-30=-50

=>-28x=-20

=>x=20/28=5/7

c: =>2x+x^3-x=x^3+1

=>x=1

d: =>x^3-3x^2+3x-1-x(x^2+2x+1)=10x-2x^2-11x-22

=>x^3-3x^2+3x-1-x^3-2x^2-x=-2x^2-x-22

=>-5x^2+2x-1+2x^2+x+22=0

=>-3x^2+3x+21=0

=>x^2-x-7=0

=>\(x=\dfrac{1\pm\sqrt{29}}{2}\)

a: Ta có: \(x\left(2x-3\right)-\left(2x-1\right)\left(x+5\right)=17\)

\(\Leftrightarrow2x^2-3x-2x^2-10x+x+5=17\)

\(\Leftrightarrow-12x=12\)

hay x=-1

tìm x nha

c, \(3x^2-7x+10=0\)

\(\Leftrightarrow3x^2+3x-10x+10=0\)

\(\Leftrightarrow3x\left(x+1\right)-10\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{10}{3}\end{matrix}\right.\)

d, \(2x\left(x-10\right)-x+10=0\)

\(\Leftrightarrow2x\left(x-10\right)-\left(x-10\right)=0\)

\(\Leftrightarrow\left(x-10\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=\dfrac{1}{2}\end{matrix}\right.\)

`#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.`

\(\dfrac{3}{2x+10}+\dfrac{3}{x-5}-\dfrac{2x}{x^2-25}\)

\(=\dfrac{3\left(x-5\right)}{2\left(x+5\right)\left(x-5\right)}+\dfrac{6\left(x+5\right)}{2\left(x+5\right)\left(x-5\right)}-\dfrac{4x}{2\left(x+5\right)\left(x-5\right)}\)

\(=\dfrac{3x-15+6x+30-4x}{2\left(x+5\right)\left(x-5\right)}\)

\(=\dfrac{5x+15}{2\left(x+5\right)\left(x-5\right)}\)

Máy bài này khá dễ chỉ cần suy nghĩ tí là làm được.

Đặt `2x^2-x=a(a>=-1/8)`

`pt<=>1/(a+1)+3/(a+3)=10/(a+7)`

`<=>(a+3)(a+7)+3(a+1)(a+7)=10(a+1)(a+3)`

`<=>a^2+10a+21+3(a^2+8a+7)=10(a^2+4a+3)`

`<=>a^2+3a^2+10a+24a+21+21=10a^2+40a+30`

`<=>4a^2+34a+42=10a^2+40a+30`

`<=>6a^2+6a-12=0`

`<=>a^2+a-2=0`

`a+b+c=0`

`=>a_1=1,a_2=-2(l)`

`a=1=>2x^2-x=1`

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

`a+b+c=0`

`=>x_1=1,x_1=-1/2`

Vậy `S={1,-1/2}`

Đặt \(2x^2-x+1=a\left(a\ge\dfrac{7}{8}\right)\)

PTTT : \(\dfrac{1}{a}+\dfrac{3}{a+2}=\dfrac{10}{a+6}\)

\(\Leftrightarrow\left(a+2\right)\left(a+6\right)+3a\left(a+6\right)=10a\left(a+2\right)\)

\(\Leftrightarrow a^2+2a+6a+12+3a^2+18a=10a^2+20a\)

\(\Leftrightarrow-6a^2+6a+12=0\)

\(\Leftrightarrow\left(a+1\right)\left(a-2\right)=0\)

\(\Leftrightarrow a=2\)

-Thay lại a = 2 ta được : \(2x^2-x-1=0\)

<=> \(\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy ...