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Bài 1.
a)
\((x-2)(2x-1)-(2x-3)(x-1)-2\\=2x^2-x-4x+2-(2x^2-2x-3x+3)-2\\=2x^2-5x+2-(2x^2-5x+3)-2\\=2x^2-5x+2-2x^2+5x-3-2\\=(2x^2-2x^2)+(-5x+5x)+(2-3-2)\\=-3\)
b)
\(x(x+3y+1)-2y(x-1)-(y+x+1)x\\=x^2+3xy+x-2xy+2y-xy-x^2-x\\=(x^2-x^2)+(3xy-2xy-xy)+(x-x)+2y\\=2y\)
Bài 2.
a)
\((14x^3+12x^2-14x):2x=(x+2)(3x-4)\\\Leftrightarrow 14x^3:2x+12x^2:2x-14x:2x=3x^2-4x+6x-8\\ \Leftrightarrow 7x^2+6x-7=3x^2+2x-8\\\Leftrightarrow (7x^2-3x^2)+(6x-2x)+(-7+8)=0\\\Leftrightarrow 4x^2+4x+1=0\\\Leftrightarrow (2x)^2+2\cdot 2x\cdot 1+1^2=0\\\Leftrightarrow (2x+1)^2=0\\\Leftrightarrow 2x+1=0\\\Leftrightarrow 2x=-1\\\Leftrightarrow x=\frac{-1}2\)
b)
\((4x-5)(6x+1)-(8x+3)(3x-4)=15\\\Leftrightarrow 24x^2+4x-30x-5-(24x^2-32x+9x-12)=15\\\Leftrightarrow 24x^2-26x-5-(24x^2-23x-12)=15\\\Leftrightarrow 24x^2-26x-5-24x^2+23x+12=15\\\Leftrightarrow -3x+7=15\\\Leftrightarrow -3x=8\\\Leftrightarrow x=\frac{-8}3\\Toru\)
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a: \(=\dfrac{6x^3+13x^2-5x}{2x+5}=\dfrac{6x^3+15x^2-2x^2-5x}{2x+5}=3x^2-x\)
b: \(=\dfrac{x^4-6x^3+12x^2-14x+3}{x^2-4x+1}\)
\(=\dfrac{x^4-4x^3+x^2-2x^3+8x^2-2x+3x^2-12x+3}{x^2-4x+1}\)
\(=x^2-2x+3\)
d: \(=\dfrac{\left(x+y\right)^2}{x+y}=x+y\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(3x\left(7x-2\right)-14x+4=0\)
\(\Leftrightarrow3x\left(7x-2\right)-2\left(7x-2\right)=0\)
\(\Leftrightarrow\left(7x-2\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x-2=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=2\\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{7}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{2}{7};\dfrac{2}{3}\right\}\)
b) ĐKXĐ: \(x\notin\left\{0;3\right\}\)
Ta có: \(\dfrac{2x+1}{x-3}+\dfrac{5-3x}{x}=\dfrac{2x^2-15}{x^2-3x}\)
\(\Leftrightarrow\dfrac{x\left(2x+1\right)}{x\left(x-3\right)}+\dfrac{\left(5-3x\right)\left(x-3\right)}{x\left(x-3\right)}=\dfrac{2x^2-15}{x\left(x-3\right)}\)
Suy ra: \(2x^2+x+5x-15-3x^2+9x-2x^2+15=0\)
\(\Leftrightarrow-3x^2+15x=0\)
\(\Leftrightarrow-3x\left(x-5\right)=0\)
mà -3<0
nên x(x-5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=5\left(nhận\right)\end{matrix}\right.\)
Vậy: S={5}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x\left(x-2\right)\left(x+2\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=\left(x^2-2x\right)\left(x+2\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=x^3+2x^2-2x^2-4x-x^3-3x^2-9x+3x^2+9x+27\)
\(=9x-4x+27=5x+27\)
\(\left(2x+7\right)\left(4x^2-14x+49\right)-2x\left(2x-1\right)\left(2x+1\right)\)
\(=\left(2x+7\right)\left(4x^2-14+49\right)-\left(4x^2-2x\right)\left(2x+1\right)\)
\(8x^3-28x+98x+28x^2-98+343-8x^3-4x^2+4x^2+2x\)
\(\left(98x-28x+2x\right)+343=72x+343\)
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a: ĐKXD: x<>0
\(\dfrac{14x^3+12x^2-14x}{2x}=\left(x+2\right)\left(3x-4\right)\)
=>\(\dfrac{2x\left(7x^2+6x-7\right)}{2x}=\left(x+2\right)\left(3x-4\right)\)
=>\(7x^2+6x-7=3x^2-4x+6x-8\)
=>\(7x^2+6x-7=3x^2+2x-8\)
=>\(4x^2+4x+1=0\)
=>\(\left(2x+1\right)^2=0\)
=>2x+1=0
=>x=-1/2(nhận)
b: \(\left(4x-5\right)\left(6x+1\right)-\left(8x+3\right)\left(3x-4\right)=15\)
=>\(24x^2+4x-30x-5-\left(24x^2-32x+9x-12\right)=15\)
=>\(24x^2-26x-5-24x^2+23x+12=15\)
=>-3x+7=15
=>-3x=8
=>\(x=-\dfrac{8}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
chứng minh rằng giá trị của các biểu thức sau không phụ thuộc vào giá trị của biến x
(2𝑥−1)(2𝑥+2)−4(𝑥−2)(𝑥−1)−14𝑥−1
2𝑥(2𝑥+2)−1(2𝑥+2)−4(𝑥−2)(𝑥−1)−14𝑥−1
2𝑥(2𝑥+2)−1(2𝑥+2)−4(𝑥−2)(𝑥−1)−14𝑥−1
4𝑥2+4𝑥−1(2𝑥+2)−4(𝑥−2)(𝑥−1)−14𝑥−1
4𝑥2+4𝑥−(2𝑥+2)−4(𝑥−2)(𝑥−1)−14𝑥−1
4𝑥2+4𝑥−2𝑥−2−4(𝑥−2)(𝑥−1)−14𝑥−1
4𝑥2+4𝑥−2𝑥−2−4(𝑥−2)(𝑥−1)−14𝑥−1
4𝑥2+2𝑥−2−4(𝑥−2)(𝑥−1)−14𝑥−1
4𝑥2+2𝑥−2−4(𝑥−2)(𝑥−1)−14𝑥−1
4𝑥2+2𝑥−2−4𝑥(𝑥−1)+8(𝑥−1)−14𝑥−1
4𝑥2+2𝑥−2−4𝑥(𝑥−1)+8(𝑥−1)−14𝑥−1
4𝑥2+2𝑥−2−4𝑥2+4𝑥+8(𝑥−1)−14𝑥−1
4𝑥2+2𝑥−2−4𝑥2+4𝑥+8(𝑥−1)−14𝑥−1
4𝑥2+2𝑥−2−4𝑥2+4𝑥+8𝑥−8−14𝑥−1
4𝑥2+2𝑥−2−4𝑥2+4𝑥+8𝑥−8−14𝑥−1
4𝑥2+2𝑥−2−4𝑥2+12𝑥−8−14𝑥−1
4𝑥2+2𝑥−2−4𝑥2+12𝑥−8−14𝑥−1
4𝑥2+2𝑥−11−4𝑥2+12𝑥−14𝑥
4𝑥2+2𝑥−11−4𝑥2+12𝑥−14𝑥
2𝑥−11+12𝑥−14𝑥
2𝑥−11+12𝑥−14𝑥
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