Tìm x biết :
1, ( x -3 ) + ( x -2 ) + ... + 11 = 11
2, x ( x + 2 ) > 0
3, ( x + 1 ) ( x + 5 ) < 0
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TT
Tìm x biết
1. 2(5x-8)-3(4x-5)=4(3x-4)+11
2. (2x+1)2-(4x-1).(x-3)-15=0
3. (3x-1).(2x-7)-(1-3x).(6x-5)=0
2
14 tháng 10 2021
1) \(\Rightarrow10x-16-12x+15=12x-16+11\)
\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)
2) \(\Rightarrow4x^2+4x+1-4x^2+13x-3-15=0\)
\(\Rightarrow17x=17\Rightarrow x=1\)
3) \(\Rightarrow\left(3x-1\right)\left(2x-7+6x-5\right)=0\)
\(\Rightarrow\left(2x-3\right)\left(3x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
14 tháng 10 2021
2: Ta có: \(\left(2x+1\right)^2-\left(4x-1\right)\left(x-3\right)-15=0\)
\(\Leftrightarrow4x^2+4x+1-4x^2+12x+x-3-15=0\)
\(\Leftrightarrow17x=17\)
hay x=1
2 tháng 12 2023
Bài 4:
1: \(\left(x-1\right)\left(x^2+x+1\right)-x^3-6x=11\)
=>\(x^3-1-x^3-6x=11\)
=>-6x-1=11
=>-6x=11+1=12
=>\(x=\dfrac{12}{-6}=-2\)
2: \(16x^2-\left(3x-4\right)^2=0\)
=>\(\left(4x\right)^2-\left(3x-4\right)^2=0\)
=>\(\left(4x-3x+4\right)\left(4x+3x-4\right)=0\)
=>(x+4)(7x-4)=0
=>\(\left[{}\begin{matrix}x+4=0\\7x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{4}{7}\end{matrix}\right.\)
3: \(x^3-x^2-3x+3=0\)
=>\(\left(x^3-x^2\right)-\left(3x-3\right)=0\)
=>\(x^2\left(x-1\right)-3\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x^2-3\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\x^2-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)
4: \(\dfrac{x-1}{x+2}=\dfrac{x+2}{x+1}\)(ĐKXĐ: \(x\notin\left\{-2;-1\right\}\))
=>\(\left(x+2\right)^2=\left(x-1\right)\left(x+1\right)\)
=>\(x^2+4x+4=x^2-1\)
=>4x+4=-1
=>4x=-5
=>\(x=-\dfrac{5}{4}\left(nhận\right)\)
5: ĐKXĐ: \(x\notin\left\{0;-1\right\}\)
\(\dfrac{1}{x}+\dfrac{2}{x+1}=0\)
=>\(\dfrac{x+1+2x}{x\left(x+1\right)}=0\)
=>3x+1=0
=>3x=-1
=>\(x=-\dfrac{1}{3}\left(nhận\right)\)
6: ĐKXĐ: \(x\notin\left\{0;3\right\}\)
\(\dfrac{9-x^2}{x}:\left(x-3\right)=1\)
=>\(\dfrac{-\left(x^2-9\right)}{x\left(x-3\right)}=1\)
=>\(\dfrac{-\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)}=1\)
=>\(\dfrac{-x-3}{x}=1\)
=>-x-3=x
=>-2x=3
=>\(x=-\dfrac{3}{2}\left(nhận\right)\)
2 tháng 12 2023
bài 5:
1: \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{12x^3y^2:6xy^2}{18xy^5:6xy^2}=\dfrac{2x^2}{3y^3}\)
2: \(\dfrac{10xy-5x^2}{2x^2-8y^2}=\dfrac{5x\cdot2y-5x\cdot x}{2\left(x^2-4y^2\right)}\)
\(=\dfrac{5x\left(2y-x\right)}{-2\left(x+2y\right)\left(2y-x\right)}=\dfrac{-5x}{2\left(x+2y\right)}\)
3: \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)
\(=\dfrac{\left(x^2-xy\right)-\left(x-y\right)}{\left(x^2+xy\right)-\left(x+y\right)}\)
\(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
4: \(\dfrac{\left(x+1\right)\left(x^2-2x+1\right)}{\left(6x^2-6\right)\left(x^3-1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)^2}{6\left(x^2-1\right)\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)}{6\left(x-1\right)\left(x+1\right)\cdot\left(x^2+x+1\right)}\)
\(=\dfrac{1}{6\left(x^2+x+1\right)}\)
5: \(\dfrac{2x^2-7x+3}{1-4x^2}\)
\(=-\dfrac{2x^2-7x+3}{4x^2-1}\)
\(=-\dfrac{2x^2-6x-x+3}{\left(2x-1\right)\left(2x+1\right)}\)
\(=-\dfrac{2x\left(x-3\right)-\left(x-3\right)}{\left(2x-1\right)\left(2x+1\right)}\)
\(=-\dfrac{\left(x-3\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{-x+3}{2x+1}\)
Bài 3:
1: \(9x^3-xy^2\)
\(=x\cdot9x^2-x\cdot y^2\)
\(=x\left(9x^2-y^2\right)\)
\(=x\left(3x-y\right)\left(3x+y\right)\)
2: \(x^2-3xy-6x+18y\)
\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)
\(=x\left(x-3y\right)-6\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-6\right)\)
3: \(x^2-3xy-6x+18y\)
\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)
\(=x\left(x-3y\right)-6\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-6\right)\)
4: \(6xy-x^2+36-9y^2\)
\(=36-\left(x^2-6xy+9y^2\right)\)
\(=36-\left(x-3y\right)^2\)
\(=\left(6-x+3y\right)\left(6+x-3y\right)\)
5: \(x^4-6x^2+5\)
\(=x^4-x^2-5x^2+5\)
\(=x^2\left(x^2-1\right)-5\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x-1\right)\left(x+1\right)\)
6: \(9x^2-6x-y^2+2y\)
\(=\left(9x^2-y^2\right)-\left(6x-2y\right)\)
\(=\left(3x-y\right)\left(3x+y\right)-2\left(3x-y\right)\)
\(=\left(3x-y\right)\left(3x+y-2\right)\)
LN
2
7 tháng 8 2021
1, \(\left(x-1\right)\left(x+2\right)-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left[x+2-\left(x-1\right)\right]=0\)
\(\Leftrightarrow3\left(x-1\right)=0\Leftrightarrow x=1\)
2, \(\left(x-2\right)^2-3\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x-2-3\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(-2x-5\right)=0\Leftrightarrow x=-\dfrac{5}{2};x=2\)
3, \(\left(5-2x\right)\left(2x+7\right)=4x^2-25=\left(2x-5\right)\left(2x+5\right)\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(5-2x\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7+2x+5\right)=0\Leftrightarrow\left(4x+12\right)\left(5-2x\right)=0\Leftrightarrow x=-3;x=\dfrac{5}{2}\)
7 tháng 8 2021
1) Ta có: \(\left(x-1\right)\left(x+2\right)-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2-x+1\right)=0\)
\(\Leftrightarrow x-1=0\)
hay x=1
2) Ta có: \(\left(x-2\right)^2-3\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2-3x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-5}{2}\end{matrix}\right.\)
8 tháng 7 2021
1) PT \(\Leftrightarrow\dfrac{x+3}{15}=\dfrac{4}{15}\) \(\Rightarrow x+3=4\) \(\Rightarrow x=1\)
Vậy ...
2) Mạnh dạn đoán đề là \(\left(2x-5\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\x-3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=3\end{matrix}\right.\)
Vậy ...
3) PT \(\Rightarrow3x-4-2x+5=3\)
\(\Rightarrow x=2\)
Vậy ...
4) PT \(\Rightarrow\left[{}\begin{matrix}2x+1=0\\\dfrac{1}{2}x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)
Vậy ...
8 tháng 7 2021
3) Ta có: \(\left(3x-4\right)-\left(2x-5\right)=3\)
\(\Leftrightarrow3x-4-2x+5=3\)
\(\Leftrightarrow x+1=3\)
hay x=2
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 8 2021
1.
\(\left(x-5\right)^2+3\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-5+3\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)
2.
\(\left(x^2-9\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 8 2021
3.
\(\left(2x+1\right)^2+\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1+x-1\right)=0\)
\(\Leftrightarrow\left(2x+1\right).3x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\2x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
4.
\(\left(x-1\right)\left(x+3\right)+\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1+x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
14 tháng 10 2021
1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)
\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)
\(\Leftrightarrow x=2\)
3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)
\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)
\(\Leftrightarrow6x=6\)
hay x=1
TT
1
HP
14 tháng 10 2021
1. x(x + 1) - x2 + 1 = 0
<=> x(x + 1) - (x2 - 1) = 0
<=> x(x + 1) - (x + 1)(x - 1) = 0
<=> (x - x + 1)(x + 1) = 0
<=> x + 1 = 0\
<=> x = -1
2. 4x(x - 2) - 6 + 3x = 0
<=> 4x(x - 2) - (3x - 6) = 0
<=> 4x(x - 2) - 3(x - 2) = 0
<=> (4x - 3)(x - 2) = 0
<=> \(\left[{}\begin{matrix}4x-3=0\\x-2=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=2\end{matrix}\right.\)
3. x(x + 2) - 3(x + 2) = 0
<=> (x - 3)(x + 2) = 0
<=> \(\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
29 tháng 11 2015
a, (x+1) + (x+3) + (x+5) +.......+(x+99) =0
\(\Rightarrow\)x+x+x+...+x(50 số hạng) + 1+3+5+...+99=0
\(\Rightarrow\)50.x+4950=0
50.x=0-4950
50.x=(-4950)
x=(-4950):50
x=(-99)
2. x( x + 2) > 0
TH1\(\orbr{\begin{cases}x>0\\x+2>0\end{cases}}\)=> \(\orbr{\begin{cases}x>0\\x>-2\end{cases}}\)
TH2 \(\orbr{\begin{cases}x< 0\\x+2< 0\end{cases}}\) => \(\orbr{\begin{cases}x< 0\\x< -2\end{cases}}\)
3, ( x + 1) ( x + 5) < 0
TH1: \(\orbr{\begin{cases}x+1>0\\x+5< 0\end{cases}}\) => \(\orbr{\begin{cases}x>-1\\x< -5\end{cases}}\)
TH2:\(\orbr{\begin{cases}x+1< 0\\x+5>0\end{cases}}\) => \(\orbr{\begin{cases}x< -1\\x>-5\end{cases}}\)
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