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5 tháng 12 2015

\(y=\frac{\frac{^x}{x^2}-x-6-x-\frac{1}{3}x^2-4x-15}{x^4}-2x^2+\frac{1}{3}x^2+11x+10b\)

\(y=\frac{-\left(5x^7-33x^6-30bx^5+x^3+18x^2+63x-3\right)}{3x^5}\)

7 tháng 10 2023

a) \(C=\left(\dfrac{x}{x^2-x-6}-\dfrac{x-1}{3x^2-4x-15}\right):\dfrac{x^4-2x^2+1}{3x^2+11x+10}\cdot\left(x^2-2x+1\right)\) (ĐK: \(x\ne-\dfrac{5}{3};x\ne3;x\ne-2;x\ne1\))

\(C=\left[\dfrac{x}{\left(x-3\right)\left(x+2\right)}-\dfrac{x-1}{\left(x-3\right)\left(3x+5\right)}\right]:\dfrac{\left(x^2-1\right)^2}{\left(3x+5\right)\left(x+2\right)}\cdot\left(x-1\right)^2\)

\(C=\left[\dfrac{x\left(3x+5\right)}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-3\right)\left(3x+5\right)\left(x+2\right)}\right]\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)

\(C=\dfrac{3x^2+5x-x^2-2x+x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)

\(C=\dfrac{2x^2+4x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)

\(C=\dfrac{2\left(x+1\right)^2}{\left(3x+5\right)\left(x-3\right)\left(x+2\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)

\(C=\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}\)

b) Thay x = 2003 ta có: 

\(C=\dfrac{2}{\left(2003-1\right)^4\left(2003-3\right)}=\dfrac{2}{2002^4\cdot2000}=\dfrac{1}{2002^4\cdot1000}\)

c) \(C>0\) khi: 

\(\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}>0\) mà: \(\left\{{}\begin{matrix}2>0\\\left(x-1\right)^4>0\end{matrix}\right.\)

\(\Leftrightarrow x-3>0\)

\(\Leftrightarrow x>3\) (đpcm) 

24 tháng 10 2023

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)

\((4x-5)(6x+1)-(8x+3)(3x-4)=15\\\Leftrightarrow 24x^2+4x-30x-5-(24x^2-32x+9x-12)=15\\\Leftrightarrow 24x^2-26x-5-(24x^2-23x-12)=15\\\Leftrightarrow 24x^2-26x-5-24x^2+23x+12=15\\\Leftrightarrow -3x+7=15\\\Leftrightarrow -3x=8\\\Leftrightarrow x=\frac{-8}3\\Toru\)

22 tháng 10 2023

1:

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

\(=4x^2-20x+25-4x^2-12x\)

=-32x+25

b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)

\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)

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

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

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

\(=9+10x-15=10x-6\)

2: 

a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)

\(=9x^2-12x+4-5x^2+20x+4x-4\)

\(=4x^2+12x\)

b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)

\(=27-x^3+x^3-9x^2+27x-27\)

\(=-9x^2+27x\)

c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)

\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)

\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)

\(=-5\left(x^2-16\right)=-5x^2+80\)

28 tháng 12 2021

Câu 1: C

Câu 2: =x(x-2)*(x+2)

5 tháng 12 2015

dễ mà, tik đi, mik giải cho!

5 tháng 12 2015

Đừng mắc lừa nhưng kẻ giả vờ rủ òng thương người khác

21 tháng 7 2021

a/ 2x\(^{^{ }3}\)-3\(^{^{ }3}\)-2x\(^3\)-1\(^{^{ }3}\)=-28

b/x\(^{^{ }3}\)+2\(^{^{ }3}\)-x\(^3\)+2=10

c/3x\(^3\)+5\(^3\)-3x(3x\(^2\)-1)=3x\(^3\)+5\(^3\)-3x\(^3\)+3x=125+3x

d/ x\(^6\)-(x\(^3\)+1)(x\(^2\)-x+1)= x\(^6\)-(x\(^6\)-x\(^4\)+x\(^3\)+x\(^2\)-x+1)=x\(^4\)-x\(^3\)-x\(^2\)+x-1