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15 tháng 11 2021
\(a,=x^2+x+4x+4=\left(x+1\right)\left(x+4\right)\\ b,=x^2+2x-3x-6=\left(x-3\right)\left(x+2\right)\\ c,=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\\ d,=3\left(x^2-2x+5x-10\right)=3\left(x-2\right)\left(x+5\right)\\ e,=-3x^2+6x-x+2=\left(x-2\right)\left(1-3x\right)\\ f,=x^2-x-6x+6=\left(x-1\right)\left(x-6\right)\\ h,=4\left(x^2-3x-6x+18\right)=4\left(x-3\right)\left(x-6\right)\\ i,=3\left(3x^2-3x-8x+5\right)=3\left(x-1\right)\left(3x-8\right)\\ k,=-\left(2x^2+x+4x+2\right)=-\left(2x+1\right)\left(x+2\right)\\ l,=x^2-2xy-5xy+10y^2=\left(x-2y\right)\left(x-5y\right)\\ m,=x^2-xy-2xy+2y^2=\left(x-y\right)\left(x-2y\right)\\ n,=x^2+xy-3xy-3y^2=\left(x+y\right)\left(x-3y\right)\)
MA
1
15 tháng 11 2021
a) ĐKXĐ: \(\left\{{}\begin{matrix}2x+3\ne0\\2x+1\ne0\\\left(2x+3\right)\left(2x+1\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{3}{2}\\x\ne-\dfrac{1}{2}\\\left(2x+3\right)\left(2x+1\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{3}{2}\\x\ne-\dfrac{1}{2}\end{matrix}\right.\)
b) \(\Rightarrow P=\dfrac{2\left(2x+1\right)+3\left(2x+3\right)-6x-5}{\left(2x+3\right)\left(2x+1\right)}\)
\(\Rightarrow P=\dfrac{4x+2+6x+9-6x-5}{\left(2x+3\right)\left(2x+1\right)}\)
\(\Rightarrow P=\dfrac{4x+6}{\left(2x+3\right)\left(2x+1\right)}\)
\(\Rightarrow P=\dfrac{2\left(2x+3\right)}{\left(2x+3\right)\left(2x+1\right)}\)
\(\Rightarrow P=\dfrac{2}{2x+1}\)
c) \(P=-1\Rightarrow\dfrac{2}{2x+1}=-1\\ \Rightarrow2=-2x-1\\ \Rightarrow2x=-3\\ \Rightarrow x=-\dfrac{3}{2}\)
MA
0
MA
4
14 tháng 11 2021
Bài 6
\(a,ĐK:x\ne\pm5\\ b,P=\dfrac{x-5+2x+10-2x-10}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}=\dfrac{1}{x+5}\\ c,P=-3\Leftrightarrow\dfrac{1}{x+5}=-3\Leftrightarrow-3\left(x+5\right)=1\Leftrightarrow x=-\dfrac{16}{3}\\ \Leftrightarrow Q=\left(3x-7\right)^2=\left[3\cdot\left(-\dfrac{16}{3}\right)-7\right]^2=529\)
14 tháng 11 2021
Bài 7:
\(a,ĐK:x\ne\pm3\\ b,P=\dfrac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}=\dfrac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{4}{x-3}\\ b,P=4\Leftrightarrow4\left(x-3\right)=4\Leftrightarrow x=4\)
HT
1
NA
38
5 tháng 5 2016
<=>x3+x3-6x2+12x-8=8x3-24x2+24x-8
<=>-6x3+18x2-12x=0
<=>-x(6x2-18x+12)=0
<=>-x(6x2-6x-12x+12)=0
<=>-x(6x-12)(x-1)=0
<=>x=0;2;1
HT
5 tháng 5 2016
Ta có \(x^3+\left(x-2\right)^3=\left(2x-2\right)^3\)
\(\Rightarrow x^3+\left(x-2\right)^3-\left(2x-2\right)^3=0\)
\(\Rightarrow x^3+\left(x-2\right)^3+\left(2-2x\right)^3=0\)
Đặt \(x=a;x-2=b;2-2x=c\)
\(a+b+c=x+x-2+2-2x=0\)
Xét bài toán phụ \(a+b+c=0\Rightarrow a^3+b^3+c^3=3abc\)
\(a^3+b^3+c^3=\left(a+b\right)^3+c^3-3a^2b-3ab^2\)
= \(\left(a+b\right)^3+c^3-3ab\left(a+b\right)\)
\(=\left(-c\right)^3+c^3-3ab\left(-c\right)=3abc\)
\(\Rightarrow x^3+\left(x-2\right)^3+\left(2-2x\right)^3=3x\left(x-2\right)\left(2-2x\right)=0\)
\(\Rightarrow x=0\) hoặc \(x-2=0\Rightarrow x=2\) hoặc \(2-2x=0\Rightarrow2x=2\Rightarrow x=1\)
Vậy phương trình có tập nghiệm \(S=\left\{0;2;1\right\}\)
16 tháng 2 2023
\(\dfrac{x}{a}=\dfrac{m-\dfrac{x}{2}}{m}\)
\(\Rightarrow xm=a\left(m-\dfrac{x}{2}\right)\)
\(\Rightarrow xm=am-\dfrac{ax}{2}\)
\(\Rightarrow2xm=2am-ax\)
\(\Rightarrow2xm+ax=2am\)
\(\Rightarrow x\left(2m+a\right)=2am\)
\(\Rightarrow x=\dfrac{2am}{a+2m}\)
VH
6 tháng 8 2017
A= m2-m+1= m2-2m.1/2 +(1/2)2-(1/2)2 +1=(m-1/2)2 +5/4 lớn hơn hoặc = 5/4
do đó A nhỏ nhất khi bằng 5/4
=> (m-1/2)2+5/4 = 5/4
=>(m-1/2)2=0
=>m-1/2=0
=> m=1/2
nếu đúng thì k cho mình nka
a.\(\dfrac{x}{3}+\dfrac{2x-1}{5}=2-\dfrac{x}{3}\)
\(\Leftrightarrow\dfrac{5x+3\left(2x-1\right)}{15}=\dfrac{30-5x}{15}\)
\(\Leftrightarrow5x+3\left(2x-1\right)=30-5x\)
\(\Leftrightarrow5x+6x-3-30+5x=0\)
\(\Leftrightarrow16x-33=0\)
\(\Leftrightarrow16x=33\)
\(\Leftrightarrow x=\dfrac{33}{16}\)
b.\(\left(2x-1\right)\left(3x+2\right)=\left(5x-8\right)\left(2x-1\right)\)
\(\Leftrightarrow\left(2x-1\right)\left(3x+2\right)-\left(5x-8\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x+2-5x+8\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(10-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=5\end{matrix}\right.\)
c.\(ĐK:x\ne-1;3\)
\(\Rightarrow\dfrac{x-1}{x+1}+\dfrac{x+5}{x-3}=\dfrac{8}{\left(x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-3\right)+\left(x+5\right)\left(x+1\right)}{\left(x+1\right)\left(x-3\right)}=\dfrac{8}{\left(x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)+\left(x+5\right)\left(x+1\right)=8\)
\(\Leftrightarrow x^2-3x-x+3+x^2+x+5x+5-8=0\)
\(\Leftrightarrow2x^2+2x=0\)
\(\Leftrightarrow2x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
a: \(\Leftrightarrow5x+3\left(2x-1\right)=30-5x\)
=>5x+6x-3=30-5x
=>11x-3=30-5x
=>16x=33
hay x=33/16
b: \(\Leftrightarrow\left(2x-1\right)\left(5x-8\right)-\left(2x-1\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(5x-8-3x-2\right)=0\)
=>(2x-1)(2x-10)=0
=>x=1/2 hoặc x=5
c: \(\Leftrightarrow\dfrac{x-1}{x+1}+\dfrac{x+5}{x-3}=\dfrac{8}{\left(x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow x^2-4x+3+x^2+6x+5=8\)
\(\Leftrightarrow2x^2+2x=0\)
=>2x(x+1)=0
=>x=0(nhận) hoặc x=-1(loại)