Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
c: \(\dfrac{3x+5}{x^2-5x}+\dfrac{25-x}{25-5x}\)
\(=\dfrac{3x+5}{x\left(x-5\right)}+\dfrac{x-25}{5\left(x-5\right)}\)
\(=\dfrac{15x+25+x^2-25x}{5x\left(x-5\right)}=\dfrac{x^2-10x+25}{5x\left(x-5\right)}=\dfrac{x-5}{5x}\)
e: \(\dfrac{4x^2-3x+17}{x^3-1}+\dfrac{2x-1}{x^2+x+1}+\dfrac{6}{1-x}\)
\(=\dfrac{4x^2-3x+17+\left(2x-1\right)\left(x-1\right)-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{-2x^2-9x+11+2x^2-3x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-12}{x^2+x+1}\)
mình chỉ phân tích thôi
a) 6x(4-x)+x-4
=6x(4-x)-(4-x)
=(6x-1)(4-x)
c) 25x^2-10x+1-16z^2
=(5x-1)^2-16z^2
=(5x-1-4z)(5x-1+4z)
ban xem lại đề bài câu b đi chắc là sai đó
còn các câu trên bạn tự làm nhé
Thực hiện phép tính:
a) (2x-3y)(4x2+6xy+9y2)
=8x3-27y3
b) (6x3+3x2+4x+2):(3x2+2)
=(3x2+2)(2x+1):(3x2+2)
=2x+1
c) (x+2)2+(3-x)-2(x+3)(x-3)
=x2+4x+4+3-x-2x2+18
=-x2+4x+25
Bài 2:
a: 6x(4-x)+(x-4)
=6x(4-x)-(4-x)
=(4-x)(6x-1)
b: \(=x^2-1+y\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(1+y\right)=\left(y+1\right)\left(x+1\right)\left(x-1\right)\)
c: \(=\left(5x-1\right)^2-\left(4z\right)^2\)
=(5x-1-4z)(5x-1+4z)
\(a.\frac{x-5}{4}-2x+1=\frac{x}{3}-\frac{2-x}{6}\\\Leftrightarrow \frac{3\left(x-5\right)}{12}-\frac{24}{12}x+\frac{12}{12}=\frac{4x}{12}-\frac{2\left(2-x\right)}{12}\\\Leftrightarrow 3\left(x-5\right)-24x+12=4x-2\left(2-x\right)\\\Leftrightarrow 3x-15-24x+12=4x-4+2x\\ \Leftrightarrow3x-15-24x+12-4x+4-2x=0\\ \Leftrightarrow-27x+1=0\\ \Leftrightarrow-27x=-1\\ \Leftrightarrow x=\frac{1}{27}\)
\(b.\left(2x-1\right)^2=\left(x-2\right)\left(2x-1\right)\\ \Leftrightarrow\left(2x-1\right)^2-\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left[\left(2x-1\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1-x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-1\end{matrix}\right.\)
\(c.\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{-3}{25-x^2}\\\Leftrightarrow \frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{3}{x^2-25}\\\Leftrightarrow \frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{3}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\frac{\left(x+5\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\frac{3}{\left(x-5\right)\left(x+5\right)}\\ \Leftrightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=3\\\Leftrightarrow x^2+5x+5x+25-\left(x^2-5x-5x+25\right)=3\\\Leftrightarrow x^2+5x+5x+25-x^2+5x+5x-25=3\\ \Leftrightarrow20x=3\\ \Leftrightarrow x=\frac{3}{20}\)
\(d.x^2-x-12=0\\\Leftrightarrow x^2-4x+3x-12=0\\\Leftrightarrow \left(x^2-4x\right)+\left(3x-12\right)=0\\ \Leftrightarrow x\left(x-4\right)+3\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
heoheo lần sau bạn đánh = kí hiệu đi :(((
a/ \(\dfrac{x}{3}+\dfrac{2x-1}{6}=\dfrac{1}{2}\)
\(\Leftrightarrow2x+2x-1=3\)
<=> 4x = 4 <=> x = 1
Vậy x = 1
b/ \(\dfrac{3x+1}{2}+\dfrac{x-1}{3}=\dfrac{x-9}{6}\)
\(\Leftrightarrow3\left(3x+1\right)+2\left(x-1\right)=x-9\)
\(\Leftrightarrow9x+3+2x-2=x-9\)
\(\Leftrightarrow10x=-10\Leftrightarrow x=-1\)
Vậy pt có nghiệm x = -1
c/ \(\dfrac{x-1}{x-2}=\dfrac{x+3}{x+2}\) ĐKXĐ: \(x\ne\pm2\)
<=> \(\left(x-1\right)\left(x+2\right)=\left(x+3\right)\left(x-2\right)\)
\(\Leftrightarrow x^2+2x-x-2=x^2-2x+3x-6\)
\(\Leftrightarrow0x=-4\left(voly\right)\)
Vậy pt vô nghiệm
d/ \(\dfrac{3x-1}{3x+1}+\dfrac{x-3}{x+3}=2\) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-3\\x\ne-\dfrac{1}{3}\end{matrix}\right.\)
pt <=> \(\dfrac{\left(3x-1\right)\left(x+3\right)}{\left(3x+1\right)\left(x+3\right)}+\dfrac{\left(x-3\right)\left(3x+1\right)}{\left(3x+1\right)\left(x+3\right)}=\dfrac{2\left(3x+1\right)\left(x+3\right)}{\left(3x+1\right)\left(x+3\right)}\)
=> (3x-1)(x+3) + (x-3)(3x+1) = 2(3x+1)(x+3)
\(\Leftrightarrow3x^2+8x-3+3x^2-8x-3=6x^2+20x+6\)
\(\Leftrightarrow-20x=12\Leftrightarrow x=-\dfrac{3}{5}\left(tm\right)\)
Vậy pt có nghiệm x=....
e/ như ý d