Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
2 câu dễ làm trước, 2 câu còn lại tối đi học về mới làm được..(giờ bận rồi)
a) ĐẶt \(x^2+3x+1=a\)
\(A=a\left(a-4\right)-5=a^2-4a-5=\left(a-5\right)\left(a+1\right)\)
\(=\left(x^2+3x-4\right)\left(x^2+3x+2\right)\)
\(=\left(x-1\right)\left(x+4\right)\left(x+1\right)\left(x+2\right)\)
c)\(C=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt ẩn phụ: \(t=x^2+8x+7\) rồi làm tiếp đi..
Để anh làm nốt vậy.
\(B=\left(x^2+2x\right)^2-2x^2-4x-3\)
\(B=\left(x^2+2x\right)^2-2\left(x^2+2x\right)+1-4\)
\(B=\left(x^2+2x-1\right)^2-2^2\)
\(B=\left(x^2+2x-3\right)\left(x^2+2x+1\right)\)
\(B=\left(x+3\right)\left(x-1\right)\left(x+1\right)^2\)
___
\(D=x^2-2xy+y^2-7x+7y+12\)
\(D=\left(x-y\right)^2-7\left(x-y\right)+12\)
\(D=\left(x-y\right)^2-3\left(x-y\right)-4\left(x-y\right)+12\)
\(D=\left(x-y\right)\left(x-y-3\right)-4\left(x-y-3\right)\)
\(D=\left(x-y-3\right)\left(x-y-4\right)\)
\(5\left(x+3\right)-2x\left(x+3\right)=0\)
<=> \(\left(5-2x\right)\left(x+3\right)=0\)
<=> \(\hept{\begin{cases}5-2x=0\\x+3=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{5}{2}\\x=-3\end{cases}}\)
\(4x\left(x-2018\right)-x+2018=0\)
<=> \(4x\left(x-2018\right)-\left(x-2018\right)=0\)
<=> \(\left(4x-1\right)\left(x-2018\right)=0\)
<=> \(\hept{\begin{cases}4x-1=0\\x-2018=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{1}{4}\\x=2018\end{cases}}\)
\(\left(x+1\right)^2-\left(x+1\right)=0\)
<=> \(\left(x+1\right)\left(x+1-1\right)=0\)
<=> \(\left(x+1\right).x=0\)
<=> \(\hept{\begin{cases}x=0\\x+1=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=0\\x=-1\end{cases}}\)
học tốt
a) \(5\left(x+3\right)-2x\left(3+x\right)=0\)
\(5\left(x+3\right)+2x\left(x+3\right)=0\)
\(\left(x+3\right)\left(5+2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\5+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{-5}{2}\end{cases}}\)
b) \(4x\left(x-2018\right)-x+2018=0\)
\(4x\left(x-2018\right)-\left(x-2018\right)=0\)
\(\left(x-2018\right)\left(4x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2018=0\\4x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2018\\x=\frac{1}{4}\end{cases}}\)
c) \(\left(x+1\right)^2-\left(x+1\right)=0\)
\(\left(x+1\right)\left(x+1-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\x+1-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}\)
\(\left(x^2+2x\right)^2-2x^2-4x=3\)
\(\Rightarrow x^4+4x^3+4x^2-2x^2-4x=3\)
\(\Rightarrow x^4+4x^3+2x^2-4x-3=0\)
\(\Rightarrow x^3\left(x-1\right)+5x^2\left(x-1\right)+7x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x^3+5x^2+7x+3\right)=0\)
\(\Rightarrow\left(x-1\right)\left[x^2\left(x+1\right)+4x\left(x+1\right)+3\left(x+1\right)\right]=0\)
\(\Rightarrow\left(x-1\right)\left(x+1\right)\left(x^2+4x+3\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+1\right)\left[x\left(x+3\right)+\left(x+3\right)\right]=0\)
\(\Rightarrow\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=-3\end{matrix}\right.\)
a) \(x^3+2x^2-4x+1\)
\(=\left(x^3+3x^2-x\right)-\left(x^2+3x-1\right)\)
\(=x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)
\(=\left(x-1\right)\left(x^2+3x-1\right)\)
c) cho da thuc P(x) =2x^4-7x^3 -2x^2 +13x +6? | Yahoo Hỏi & Đáp
Tham khảo
\(\frac{1}{x+1}+\frac{1}{x-1}+\frac{2}{1-x^2}\)
\(=\frac{1}{x+1}+\frac{1}{x-1}-\frac{2}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x-1+x+1-2}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2x-2}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2}{x+1}\)
a: \(\Leftrightarrow\left(4x+12\right)\left(3x-2\right)-\left(3x+3\right)\left(4x-1\right)=-27\)
\(\Leftrightarrow12x^2-8x+36x-24-\left(12x^2-3x+12x-3\right)=-27\)
\(\Leftrightarrow12x^2+28x-24-12x^2-9x+3=-27\)
\(\Leftrightarrow19x-21=-27\)
=>19x=-6
hay x=-6/19
b: \(\left(x+1\right)\left(3x^2-x+1\right)+x^2\left(4-3x\right)=\dfrac{5}{2}\)
\(\Leftrightarrow3x^3-x^2+x+3x^2-x+1+4x^2-3x^3=\dfrac{5}{2}\)
\(\Leftrightarrow6x^2+1=\dfrac{5}{2}\)
\(\Leftrightarrow6x^2=\dfrac{3}{2}\)
\(\Leftrightarrow x^2=\dfrac{3}{12}=\dfrac{1}{4}\)
=>x=1/2 hoặc x=-1/2
c: \(\Leftrightarrow2\left(x^2-4\right)-4\left(x^2-x-2\right)+\left(5x+8\right)\left(x+2\right)=0\)
\(\Leftrightarrow2x^2-8-4x^2+4x+8+5x^2+10x+8x+16=0\)
\(\Leftrightarrow3x^2+22x+16=0\)
\(\text{Δ}=22^2-4\cdot3\cdot16=292>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-22-2\sqrt{73}}{6}=\dfrac{-11-\sqrt{73}}{3}\\x_2=\dfrac{-11+\sqrt{73}}{3}\end{matrix}\right.\)
d: \(\Leftrightarrow20x^2-16x-1=10x^2-2x+5x-1\)
\(\Leftrightarrow10x^2-19x=0\)
=>x(10x-19)=0
=>x=0 hoặc x=19/10
đặt ẩn phụ là ra
tích cho t đi
a) (2018x - 1) - 2019x (2018x - 1)=0
<=> (2018x - 1)(1 - 2019x)=0
<=> 2018x-1=0
1-2019x=0
<=> x=1/2018
x=1/2019