![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Mình ko ghi lại đề , bạn ghi ra xong rồi suy ra như mình nha .
1) \(=>A=\left(6x^2+3x-10x-5\right)-\left(6x^2+14x-9x-21\right)\)
\(=>A=-12x+16\)
2) \(=>B=8x^3+27-8x^3+2=29\)
3)\(=>C=[\left(x-1\right)-\left(x+1\right)]^3=\left(-2\right)^3=-8\)
4)\(=>D=[\left(2x+5\right)-\left(2x\right)]^3=5^3=125\)
5)\(=>E=\left(3x+1\right)^2-\left(3x+5\right)^2+12x+2\left(6x+3\right)\)
\(=>E=\left(3x+1+3x+5\right)\left(3x+1-3x-5\right)+12x+12x+6\)
\(=>E=\left(6x+6\right)\left(-4\right)+24x+6=-24x-24+24x+6=-18\)
6)\(=>F=\left(2x^2+3x-10x-15\right)-\left(2x^2-6x\right)+x+7=-8\)
k cho mik nha ,
![](https://rs.olm.vn/images/avt/0.png?1311)
2:
=>x^3-1-2x^3-4x^6+4x^6+4x=6
=>-x^3+4x-7=0
=>x=-2,59
4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50
=>-62x+12=-50
=>x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: x khác +-1
\(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)
<=> \(\frac{6}{\left(x-1\right)\left(x+1\right)}+5=\frac{8x-1}{4\left(x+1\right)}-\frac{12x-1}{4\left(1-x\right)}\)
<=> 24(1 - x) + 20(x + 1)(x - 1)(1 - x) = (8x - 1)(x - 1)(1 - x) - (12x - 1)(x + 1)(x - 1)
<=> 4 - 4x + 20x^2 = 18x^2 + 2x
<=> 4 - 4x + 20x^2 - 18x^2 + 2x = 0
<=> 4 - 6x + 2x^2 = 0
<=> 2(2 - 3x + x^2) = 0
<=> 2(x - 1)(x - 2) = 0
<=> x - 1 = 0 hoặc x - 2 = 0
<=> x = 1 (ktm) hoặc x = 2 (tm)
=> x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐK: x khác 1; - 1
\(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}.\)
<=> \(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}+\frac{12x-1}{4x-4}.\)
<=> \(\frac{6.4}{4\left(x^2-1\right)}+\frac{5\left(x^2-1\right)}{4\left(x^2-1\right)}=\frac{\left(8x-1\right)\left(x-1\right)}{4\left(x^2-1\right)}+\frac{\left(12x-1\right)\left(x+1\right)}{4\left(x^2-1\right)}.\)
<=> \(24+20x^2-20=8x^2-x-8x+1+12x^2-x+12x-1\)
<=> \(2x=4\)
<=> x = 2 thỏa mãn.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐKXĐ:x\ne\pm1\)
\(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)
\(\Leftrightarrow\frac{6}{\left(x-1\right)\left(x+1\right)}+5-\frac{8x-1}{4\left(x+1\right)}-\frac{12x-1}{4\left(x-1\right)}=0\)
\(\Leftrightarrow\frac{24+20\left(x^2-1\right)-\left(8x-1\right)\left(x-1\right)-\left(12x-1\right)\left(x+1\right)}{4\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow24+20x^2-20-8x^2+9x-1-12x^2-11x+1=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow x=2\)
Vậy tập nghiệm của phương trình là \(S=\left\{2\right\}\)
ĐKXĐ: \(x\ne\pm1\)
\(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)
\(\Leftrightarrow\frac{6}{\left(x+1\right)\left(x-1\right)}+5=\frac{8x-1}{4\left(x+1\right)}-\frac{12x-1}{4\left(1-x\right)}\)
\(\Leftrightarrow24\left(1-x\right)+20\left(x+1\right)\left(x-1\right)\left(1-x\right)=\left(8x-1\right)\left(x-1\right)\left(1-x\right)\)\(-\left(12x-1\right)\left(x+1\right)\left(1-x\right)\)
\(\Leftrightarrow4-4x+20x^2-20x^3=18x^2-20x^3+2x\)
\(\Leftrightarrow4-4x+20x^2=18x^2+2x\)
\(\Leftrightarrow4-4x+20x^2-18x^2-2x=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(\Leftrightarrow x^2-25-x^2-8x-16+\left(4x+1\right)^3=64x^3+8+48x^2-12x\)
\(\Leftrightarrow-8x-41+64x^3+48x^2+12x+1=64x^3+48x^2-12x+8\)
=>4x-40=-12x+8
=>16x=48
hay x=3
2: \(\Leftrightarrow12x^2-48x-x^3+1+x^3-12x^2+48x-64=x^2-2x-3-x^2-10x-25\)
\(\Leftrightarrow-63=-12x-28\)
=>12x+28=63
=>12x=35
hay x=35/12