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24 tháng 7 2018

a) \(3x^2-3y^2-x-y\)

\(\Leftrightarrow3\left(x^2-y^2\right)-x-y\)

\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)

\(\Leftrightarrow3\left(x-y\right)\)

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

\(\Leftrightarrow3x^2-7x+7-3\)

\(\Leftrightarrow\left(3x^2-3\right)-\left(7x-7\right)\)

\(\Leftrightarrow3\left(x^2-1\right)-7\left(x-1\right)\)

\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(3\left(x+1\right)-7\right)\)

\(\Leftrightarrow\left(x+1\right)\left(3x-6\right)\)

e) \(-2x^2+3x-1\)

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

\(\Leftrightarrow\left(-2x-1\right)\left(-2x+1\right)+3x\)

f) \(x^2+2xy+y^2-2x-2y\)

\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)

\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)

k) \(2x^2+5x+3\)

\(\Leftrightarrow2x^2+2x+3x+3\)

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)\)

\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)\)

l) \(x^2-2x-y^2+1\)

\(\Leftrightarrow\left(x^2-2x+1\right)-y^2\)

\(\Leftrightarrow\left(x-1\right)^2-y^2\)

\(\Leftrightarrow\left(x-1-y\right)\left(x-1+y\right)\)

24 tháng 7 2018

a) \(3x^2-3y^2-x-y\)

\(\Leftrightarrow3\left(x^2-y^2\right)-x-y\)

\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)

\(\Leftrightarrow3\left(x-y\right)\)

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

\(\Leftrightarrow3x^2-7x+7-3\)

\(\Leftrightarrow\left(3x^2-3\right)-\left(7x-7\right)\)

\(\Leftrightarrow3\left(x^2-1\right)-7\left(x-1\right)\)

\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(3\left(x+1\right)-7\right)\)

\(\Leftrightarrow\left(x+1\right)\left(3x-6\right)\)

e) \(-2x^2+3x-1\)

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

\(\Leftrightarrow\left(-2x-1\right)\left(-2x+1\right)+3x\)

f) \(x^2+2xy+y^2-2x-2y\)

\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)

\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)

k) \(2x^2+5x+3\)

\(\Leftrightarrow2x^2+2x+3x+3\)

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)\)

\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)\)

l) \(x^2-2x-y^2+1\)

\(\Leftrightarrow\left(x^2-2x+1\right)-y^2\)

\(\Leftrightarrow\left(x-1\right)^2-y^2\)

\(\Leftrightarrow\left(x-1-y\right)\left(x-1+y\right)\)

Đề bài là gì sao không ghi rõ?? 

Bài 4:

a: \(=7xy\left(2-3-4\right)=-35xy\)

b: \(=\left(x-5\right)\left(x+y\right)\)

c: \(=10x\left(x-y\right)+8\left(x-y\right)=2\left(x-y\right)\left(5x+4\right)\)

d: \(=\left(x+y\right)^3-\left(x+y\right)\)

=(x+y)(x+y+1)(x+y-1)

e: =x^2+8x-x-8

=(x+8)(x-1)

f: \(=2x^2-4x+x-2=\left(x-2\right)\left(2x+1\right)\)

g: =-5x^2+15x+x-3

=(x-3)(-5x+1)

h: =x^2-3xy+xy-3y^2

=x(x-3y)+y(x-3y)

=(x-3y)*(x+y)

Bài 4:

a: \(=7xy\left(2-3-4\right)=-35xy\)

b: \(=\left(x-5\right)\left(x+y\right)\)

c: \(=10x\left(x-y\right)+8\left(x-y\right)=2\left(x-y\right)\left(5x+4\right)\)

d: \(=\left(x+y\right)^3-\left(x+y\right)\)

=(x+y)(x+y+1)(x+y-1)

e: =x^2+8x-x-8

=(x+8)(x-1)

f: \(=2x^2-4x+x-2=\left(x-2\right)\left(2x+1\right)\)

g: =-5x^2+15x+x-3

=(x-3)(-5x+1)

h: =x^2-3xy+xy-3y^2

=x(x-3y)+y(x-3y)

=(x-3y)*(x+y)

21 tháng 10 2021

\(2x\left(x^2-7x-3\right)=2x^3-14x-6x\)

\(4xy^2\left(-2x^3+y^2-7xy\right)=-8x^4y^2+4xy^5-28x^2y^3\)

21 tháng 10 2021

all ạ

11 tháng 11 2018

a) \(x^3-4x=0\)

\(x\left(x^2-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}}\)

b) \(5x\left(3x-2\right)=4-9x^2\)

\(5x\left(3x-2\right)-\left(4-9x^2\right)=0\)

\(5x\left(3x-2\right)-\left(2-3x\right)\left(2+3x\right)=0\)

\(5x\left(3x-2\right)+\left(3x-2\right)\left(2+3x\right)=0\)

\(\left(3x-2\right)\left(5x+3x+2\right)=0\)

\(\left(3x-2\right)\left(8x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-2=0\\8x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{-1}{4}\end{cases}}}\)

c) \(x^2+7x=8\)

\(x^2+7x-8=0\)

\(x^2+8x-x-8=0\)

\(x\left(x+8\right)-\left(x+8\right)=0\)

\(\left(x+8\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+8=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-8\\x=1\end{cases}}}\)

d) \(2x^2+4y^2+10x+4xy=-25\)

\(x^2+x^2+4y^2+10x+4xy+25=0\)

\(\left(4y^2+4xy+x^2\right)+\left(x^2+10x+25\right)=0\)

\(\left(2y+x\right)^2+\left(x+5\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}2y+x=0\\x+5=0\end{cases}\Rightarrow\hept{\begin{cases}y=\frac{5}{2}\\x=-5\end{cases}}}\)

\(\left(4-3x\right)\left(10x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)

\(\left(7-2x\right)\left(4+8x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)

rồi thực hiện đến hết ... 

Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>

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

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

\(2x^2-7x+3-4x^2-4x+3=0\)

\(-2x^2-11x+6=0\)

\(2x^2+11x-6=0\)

\(2x^2+12x-x-6=0\)

\(2x\left(x+6\right)-\left(x+6\right)=0\)

\(\left(x+6\right)\left(2x-1\right)=0\)

\(x+6=0\Leftrightarrow x=-6\)

\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)

\(3x-2x^2=0\)

\(x\left(2x-3\right)=0\)

\(x=0\)

\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

Tự lm tiếp nha 

29 tháng 10 2016

Câu a nhé: 2x . x^2 - 2x . 7x - 2x . 3 = 2x^3 - 14x^2 - 6x

24 tháng 7 2020

(3x^3-2x^2+x+2)*(5x^2)

5 tháng 11 2017

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

30 tháng 9 2018

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^