\(2x+5=3^4-3^2 \)

⇔\(2x+5=9\)

⇔\(2x=4\)

⇔\(x=2\)

Đề bài đâu rồi ạ, có đề mới giải được ạ

Bài 1

1.\(x\left(x+3\right)\)

\(=x^2+3x\)

2.\(3x\left(x+2\right)\)

\(=3x^2+6x\)

3,\(x^2\left(3x-1\right)\)

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

4.\(-5x^3\left(3x^2-7\right)\)

\(=-15x^5+35x^3\)

5.\(3x\left(5x^2-2x-1\right)\)

\(=15x^3-6x^2-3x\)

6.\(-x^2\left(5x^3-x-\dfrac{1}{2}\right)\)

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

7.\(\left(x^2+2x-3\right).\left(-x\right)\)

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

8.\(4x^3\left(-2x^2+4x^4-3\right)\)

\(=-8x^5+16x^7-12x^3\)

9.\(-5x^2\left(3x^2-2x+1\right)\)

\(=-15x^4+10x^3-5x^2\)

10.\(-4x^5\left(x^3-4x^2+7x-3\right)\)

\(=-4x^8+16x^7-28x^6+12x^5\)

11.\(\left(x+2\right)\left(x+3\right)\)

\(=x^2+3x+2x+6\)

12.\(\left(x-7\right)\left(x-5\right)\)

\(=x^2-5x-7x+35\)

13.\(\left(3x+5\right)\left(2x-7\right)\)

\(=6x^2-21x+10x-35\)

14.\(\left(x-3\right)\left(x^2-2x-1\right)\)

\(x^3-2x^2-x-3x^2+6x+3\)

15.\(\left(2x-1\right)\left(x^2-5x+3\right)\)

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

16.\(\left(x-5\right)\left(-x^2+x-1\right)\)

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

17,\(\left(\dfrac{1}{2}x+3\right)\left(2x^2-4x-6\right)\)

\(=x^3-2x^2-3x+6x^2-12x-18\)

P/s:mình làm hơi tắt tại bài dài quá:))

Bài 2

1.\(5x^2-3x\left(x +2\right)\)

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

\(=2x^2+2\)

2.\(-2x^2+3\left(x^2+2\right)\)

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

\(=x^2+6\)

3.\(-4x^2+2x-4x\left(x-5\right)\)

\(=-4x^2+2x-4x^2+20x\)

\(=22x\)

4.\(3x\left(x-5\right)-5x\left(x+7\right)\)

\(=3x^2-15x-5x^2-35x\)

\(=-2x^2-50x\)

5.\(-3x\left(3x-4\right)+2x\left(3x+1\right)\)

\(=-9x^2+12x+6x^2+2x\)

\(=-3x^2+14x\)

6.\(x\left(x^2+x+1\right)-x\left(2x^2+1\right)\)

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

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

7.\(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)

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

\(=-3x^3-3x\)

8.\(4x\left(x^2-x+1\right)-x\left(3x-2x-5\right)\)

\(=4x^3-4x^2+4x-3x^2+2x^2+5x\)

\(=4x^3-5x^2+9x\)

9.\(4x\left(x^2-x+1\right)+\left(x-1\right)\left(x^2-x\right)\)

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

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

10.\(\left(2x-3\right)\left(x+4\right)+\left(-x+1\right)\left(x-2\right)\)

\(=2x^2+8x-3x-12+-x^2+2x+x-2\)

\(=x^2+8x-14\)

11.\(\left(-x+5\right)\left(x+3\right)+\left(2x-1\right)\left(x+3\right)\)

\(=-x^2-3x+5x+15+2x^2+6x-x-3\)

\(=x^2+7x+12\)

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

\(=x^2-x+3x-3-4x^3+8x^2\)

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

Bài nhiều câu như thế này lần sau bạn lưu ý tách ra các post riêng nhé.

a: \(=2x^2-3x+1+3x^2+2x-1=5x^2-x\)

b: \(=4x^3-2x^2+3x-2x^3-3x^2+4x=2x^3-5x^2+7x\)

c: \(=x^2-5x+6-3x^2+2x-1=-2x^2-3x+5\)

d: \(=2x^3+5x^2-3x+1-x^3+2x^2-x+1\)

\(=x^3+7x^2-4x+2\)

e: \(=3x^2+2x-4+4x^2-x+5=7x^2+x+1\)

f: \(=x^3-2x^2+5x-1-2x^3-3x^2+4x-2=-x^3-5x^2+9x-3\)

g: \(=4x^4-3x^3+x^2+2x-1+2x^3-4x^2+3x-1\)

\(=4x^4-x^3-3x^2+5x-2\)

ĐK: `x \ne kπ`

`cot(x-π/4)+cot(π/2-x)=0`

`<=>cot(x-π/4)=-cot(π/2-x)`

`<=>cot(x-π/4)=cot(x-π/2)`

`<=> x-π/4=x-π/2+kπ`

`<=>0x=-π/4+kπ` (VN)

Vậy PTVN.

hahihihihi

\(1,ĐK:x\ge2\\ PT\Leftrightarrow\sqrt{3x-6}+x-2-\left(\sqrt{2x-3}-1\right)=0\\ \Leftrightarrow\dfrac{3\left(x-2\right)}{\sqrt{3x-6}}+\left(x-2\right)-\dfrac{2\left(x-2\right)}{\sqrt{2x-3}+1}=0\\ \Leftrightarrow\left(x-2\right)\left(\dfrac{3}{\sqrt{3x-6}}-\dfrac{2}{\sqrt{2x-3}+1}+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\\dfrac{3}{\sqrt{3x-6}}-\dfrac{2}{\sqrt{2x-3}+1}+1=0\left(1\right)\end{matrix}\right.\)

Với \(x>2\Leftrightarrow-\dfrac{2}{\sqrt{2x-3}+1}>-\dfrac{2}{1+1}=-1\left(3x-6\ne0\right)\)

\(\Leftrightarrow\left(1\right)>0-1+1=0\left(vn\right)\)

Vậy \(x=2\)

\(2,ĐK:x\ge-1\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x^2-x+1}=b\end{matrix}\right.\left(a,b\ge0\right)\Leftrightarrow a^2+b^2=x^2+2\)

\(PT\Leftrightarrow2a^2+2b^2-5ab=0\\ \Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=2b\\b=2a\end{matrix}\right.\)

Với \(a=2b\Leftrightarrow x+1=4x^2-4x+4\left(vn\right)\)

Với \(b=2a\Leftrightarrow4x+4=x^2-x+1\Leftrightarrow x^2-5x-3=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{37}}{2}\left(tm\right)\\x=\dfrac{5-\sqrt{37}}{2}\left(tm\right)\end{matrix}\right.\)

Vậy ...

\(3,ĐK:x\ge-1\\ PT\Leftrightarrow3\left(x^2-x+1\right)-2\left(x+1\right)=5\sqrt{x^3+1}\) 

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x^2-x+1}=b\end{matrix}\right.\left(a,b\ge0\right)\)

\(PT\Leftrightarrow3b^2-2a^2=5ab\\ \Leftrightarrow2a^2+5ab-3b^2=0\\ \Leftrightarrow\left[{}\begin{matrix}a=2b\\a=-3b\left(vn\right)\end{matrix}\right.\Leftrightarrow a=2b\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{37}}{2}\\x=\dfrac{5-\sqrt{37}}{2}\end{matrix}\right.\left(\text{giống bài 2}\right)\)

E tk nha:

45x-38x=1505

=>7x=1505

=>x=215

X*(45-38)=1505

X*7=1505

X=1505:7

X=215

ủa sao m.n toàn giải hộ tui ko vậy,tui bảo nêu cách giải mà 😑