![](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)
\(P(x) = 2x + 4{x^3} + 7{x^2} - 10x + 5{x^3} - 8{x^2}\)
\(=(4{x^3}+5{x^3})+( 7{x^2}- 8{x^2})+(2x-10x)\)
\( = 9{x^3} - {x^2} - 8x\)
Ta thấy số mũ cao nhất của biến x là 3 nên \(P(x)\) có bậc là 3
Hệ số của \({x^3}\) là 9
Hệ số của \({x^2}\)là -1
Hệ số của x là -8
Hệ số tự do là 0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(|x+1|+|x+2|+|x+3|+|x+4|=10x\)
Vì \(|x+1|>0\Rightarrow|x+1|=x+1\left(1\right)\)
\(|x+2|>0\Rightarrow|x+2|=x+2\left(2\right)\)
\(|x+3|>0\Rightarrow|x+3|=x+3\left(3\right)\)
\(|x+4|>0\Rightarrow|x+4|=x+4\left(4\right)\)
Từ \(\left(1\right);\left(2\right);\left(3\right);\left(4\right)\Rightarrow x+1+x+2+x+3+x+4=10x\)
\(\Leftrightarrow4x+10=10x\)
\(\Leftrightarrow6x=10\)
\(\Leftrightarrow x=\frac{5}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) \(-5\left(x^2-3x+1\right)+x\left(1+5x\right)=x-2\)
\(\Rightarrow-5x^2+15x-5+x+5x^2=x-2\)
\(\Rightarrow16x-5=x-2\)
\(\Rightarrow16x-x=5-2\)
\(\Rightarrow15x=3\)
\(\Rightarrow x=\dfrac{15}{3}=5\)
b) \(12x^2-4x\left(3x+5\right)=10x-17\)
\(\Rightarrow12x^2-12x^2-20x=10x-17\)
\(\Rightarrow-20x=10x-17\)
\(\Rightarrow-20x-10x=-17\)
\(\Rightarrow-30x=-17\)
\(\Rightarrow x=\dfrac{-30}{-17}=\dfrac{30}{17}\)
c) \(-4x\left(x-5\right)+7x\left(x-4\right)-3x^2=12\)
\(\Rightarrow-4x^2+20x+7x^2-28x-3x^2=12\)
\(\Rightarrow-8x=12\)
\(\Rightarrow x=\dfrac{12}{-8}=-\dfrac{4}{3}\)
Bài 2:
a) \(\left(x+5\right)\left(x-7\right)-7x\left(x-3\right)\)
\(=x^2-7x+5x-35-7x^2+21x\)
\(=-6x^2+19x-35\)
b) \(x\left(x^2-x-2\right)-\left(x-5\right)\left(x+1\right)\)
\(=x^3-x^2-2x-x^2+x-5x-5\)
\(=x^3-2x^2-6x-5\)
c) \(\left(x-5\right)\left(x-7\right)-\left(x+4\right)\left(x-3\right)\)
\(=x^2-7x-5x+35-x^2-3x+4x-12\)
\(=11x+23\)
d) \(\left(x-1\right)\left(x-2\right)-\left(x+5\right)\left(x+2\right)\)
\(=x^2-2x-x+2-x^2+2x+5x+10\)
\(=4x+12\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{F\left(x\right)}{G\left(x\right)}=\dfrac{12x^4+10x^3-x-3}{3x^2+x+1}\)
\(=\dfrac{12x^4+4x^3+4x^2+6x^3+2x^2+2x-6x^2-2x-2-x-1}{3x^2+x+1}\)
\(=4x^2+2x-2+\dfrac{-x-1}{3x^2+x+1}\)
=>Thương là 4x^2+2x-2
![](https://rs.olm.vn/images/avt/0.png?1311)
a/Ta có: M(x)+N(x) = (2x5 - 4x3 + 2x2 + 10x - 1) + (-2x5 + 2x4 + 4x3 + x2 + x - 10)
= 2x5 - 2x5 - 4x3 + 4x3 + 2x4 + 2x2 + x2 + 10x + x -1 - 10
= 2x4 + 3x2 + 11x - 11
b/ Ta có: A(x) = N(x)-M(x) = (-2x5 + 2x4 + 4x3 + x2 + x - 10) - (2x5 - 4x3 + 2x2 + 10x - 1)
= -2x5 - 2x5 + 2x4 + 4x3 + 4x3 + x2 - 2x2 + x - 10x -10 + 1
= -2x5 + 2x4 + 8x3 - x2 - 9x -9
![](https://rs.olm.vn/images/avt/0.png?1311)
- \(4x_1=6x_2=10x_3=12x_4\Leftrightarrow2x_1=3x_2=5x_3=6x_4\Leftrightarrow\frac{x_1}{\frac{1}{2}}=\frac{x_2}{\frac{1}{3}}=\frac{x_3}{\frac{1}{5}}=\frac{x_4}{\frac{1}{6}}=P\)
- Thay vào \(x_1+x_2+x_3+x_4=36\Leftrightarrow\frac{1}{2}P+\frac{1}{3}P+\frac{1}{5}P+\frac{1}{6}P=36\)
- \(\Leftrightarrow\frac{6}{5}P=36\Leftrightarrow P=30\)
- Vậy, \(x_1=\frac{1}{2}P=15;x_2=\frac{1}{3}P=10;x_3=\frac{1}{5}P=6;x_4=\frac{1}{6}P=5\)
4x1 = 6x2 = 10x3 = 12x4 => \(\frac{x_1}{\frac{1}{4}}=\frac{x_2}{\frac{1}{6}}=\frac{x_3}{\frac{1}{10}}=\frac{x_4}{\frac{1}{12}}=\frac{x_1+x_2+x_3+x_4}{\frac{1}{4}+\frac{1}{6}+\frac{1}{10}+\frac{1}{12}}=\frac{36}{\frac{36}{60}}=60\)
=> (x1 ; x2 ; x3 ; x4) = ( \(\frac{60}{4};\frac{60}{6};\frac{60}{10};\frac{60}{12}\)) = ( 15 ; 10 ; 6 ; 5 )
Ta có :\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|\left(1\right)\)
+ Điều kiện : \(4x\ge0\Leftrightarrow x\ge0\)
\(\Rightarrow\left|x+1\right|=x+1\)
\(\left|x+2\right|=x+2\)
\(\left|x+3\right|=x+3\)
\(\left|x+4\right|=x+4\)
\(\left(1\right)\Leftrightarrow x+1+x+2+x+3+x+4=10x\)
\(\Rightarrow\left(x+x+x+x\right)+\left(1+2+3+4\right)=10x\)
\(\Rightarrow4x+10=10x\)
\(\Rightarrow6x=10\)
\(\Rightarrow x=\frac{5}{3}\)
Vậy \(x=\frac{5}{3}\)
Chúc bạn học tốt !!!