A
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
H9
HT.Phong (9A5)
CTVHS
24 tháng 6 2023
1, \(A=5x\left(x^2-3\right)+x^2\left(7-5x\right)-7x^2\)
\(A=5x^3-15x+7x^2-5x^3-7x^2\)
\(A=\left(5x^3-5x^3\right)+\left(7x^2-7x^2\right)-15x\)
\(A=-15x\)
Thay \(x=-5\) vào A ta được:
\(-15\cdot-5=75\)
Vậy: ....
2. \(B=x\left(x^2-3\right)+x^2\left(7-5x\right)-7x^2\)
\(B=x^3-3x+7x^2-5x^3-7x^2\)
\(B=\left(x^3-5x^3\right)+\left(7x^2-7x^2\right)-3x\)
\(B=-4x^3-3x\)
Thay \(x=10,y=-1\) vào B ta được:
\(-4\cdot10^3-3\cdot10=-4\cdot1000-3\cdot10=-4000-30=-4030\)
Vậy: ....
20 tháng 8 2021
a: Ta có: \(\left(7x+4\right)^2-\left(7x-4\right)\left(7x+4\right)\)
\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)
\(=8\left(7x+4\right)\)
=56x+32
b: Ta có: \(8\left(x-2\right)^2-3\left(x^2-4x-5\right)-5x^2\)
\(=8x^2-32x+32-3x^2+12x+15-5x^2\)
\(=-20x+47\)
c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(x+1\right)\)
\(=x^3+3x^2+3x+1-x^3+1-3x^2-3x\)
=2
10 tháng 7 2023
\(a,\left(x+2\right)^2-9=0\\ \Leftrightarrow\left(x+2-3\right)\left(x+2+3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{1;-5\right\}\)
\(b,x^2-2x+1=25\\ \Leftrightarrow\left(x-1\right)^2=25\\ \Leftrightarrow\left(x-1\right)^2-25=0\\ \Leftrightarrow\left(x-1-5\right)\left(x-1+5\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{6;-4\right\}\)
\(c,\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\\ \Leftrightarrow25x^2+10x+1-25x^2+9=30\\ \Leftrightarrow25x^2+10x-25x^2=30-1-9\\ \Leftrightarrow10x=20\\ \Leftrightarrow x=2\\ Vậy\dfrac{ }{ }S=\left\{2\right\}\)
\(d,\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\\ \Leftrightarrow x^3-1-x\left(x^2-4\right)=5\\ \Leftrightarrow x^3-1-x^3+4x=5\\ \Leftrightarrow x^3-x^3+4x=5+1\\ \Leftrightarrow4x=6\\ \Leftrightarrow x=\dfrac{3}{2}\\ Vậy\dfrac{ }{ }S=\left\{\dfrac{3}{2}\right\}\)
10 tháng 7 2023
a: =>(x+2-3)(x+2+3)=0
=>(x-1)(x+5)=0
=>x=1 hoặc x=-5
b: =>(x-1)^2=25
=>x-1=5 hoặc x-1=-5
=>x=-4 hoặc x=6
c: =>25x^2+10x+1-25x^2+9=30
=>10x+10=30
=>x+1=3
=>x=2
d: =>x^3-1-x(x^2-4)=5
=>x^3-1-x^3+4x=5
=>4x=6
=>x=3/2
16 tháng 12 2022
1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
2
16 tháng 12 2022
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
___________________________________________________
`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
___________________________________________________
`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
___________________________________________________
`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
___________________________________________________
`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
___________________________________________________
`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
21 tháng 4 2021
Mấy ý này bản chất ko khác nhau nhé, mình làm mẫu, bạn làm tương tự mấy ý kia nhé
a, \(\left|5x\right|=x+2\)
Với \(x\ge0\)thì \(5x=x+2\Leftrightarrow x=\dfrac{1}{2}\)
Với \(x< 0\)thì \(5x=-x-2\Leftrightarrow6x=-2\Leftrightarrow x=-\dfrac{1}{3}\)
b, \(\left|7x-3\right|-2x+6=0\Leftrightarrow\left|7x-3\right|=2x-6\)
Với \(x\ge\dfrac{3}{7}\)thì \(7x-3=2x-6\Leftrightarrow5x=-3\Leftrightarrow x=-\dfrac{3}{5}\)( ktm )
Với \(x< \dfrac{3}{7}\)thì \(7x-3=-2x+6\Leftrightarrow9x=9\Leftrightarrow x=1\)( ktm )
Vậy phương trình vô nghiệm
MH
18 tháng 3 2022
\(1\text{)}\left(x-5\right)^2+3\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-5+3\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
\(2\text{)}\dfrac{2x-1}{3}-\dfrac{5x+2}{7}=x+13\)
\(\Leftrightarrow\dfrac{7\left(2x-1\right)-3\left(5x+2\right)}{21}=\dfrac{21\left(x+13\right)}{21}\)
\(\Leftrightarrow14x-7-15x-6=21x+273\)
\(\Leftrightarrow-x-13=21x+273\)
\(\Leftrightarrow-22x=286\)
\(\Rightarrow x=-\dfrac{286}{22}=-\dfrac{143}{11}\)
\(3\text{)}\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{7x-6}{4-x^2}\left(đk:x\ne\pm2\right)\)
\(\Leftrightarrow\dfrac{x-1}{2+x}+\dfrac{x}{2-x}=\dfrac{7x-6}{4-x^2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(2-x\right)+x\left(2+x\right)}{4-x^2}=\dfrac{7x-6}{4-x^2}\)
\(\Leftrightarrow2x-x^2-2+x+2x+x^2=7x-6\)
\(\Leftrightarrow x-2=7x-6\)
\(\Leftrightarrow-6x=-4\)
\(\Rightarrow x=\dfrac{2}{3}\)
29 tháng 8 2018
\(A=x^2-4x-x\left(x-4\right)-15\)
\(=x^2-4x-x^2+4x-15=-15\) => đpcm
\(B=5x\left(x^2-x\right)-x^2\left(5x-5\right)-13\)
\(=5x^3-5x^2-5x^3+5x^2-13=-13\) => đpcm
\(C=-3x\left(x-5\right)+3\left(x^2-4x\right)-3x+7\)
\(=-3x^2+15x+3x^2-12x-3x+7=7\) => đpcm
29 tháng 8 2018
\(D=7\left(x^2-5x+3\right)-x\left(7x-35\right)-14\)
\(=7x^2-35x+21-7x^2+35x-14=7\) => đpcm
\(E=4x\left(x^2-7+2\right)-4\left(x^3-7x+2x-5\right)\)
\(=4x^3-20x-4x^3+20x+20=20\) => đpcm
\(H=x\left(5x-3\right)-x^2\left(x-1\right)+x\left(x^2-6x\right)-10+3x\)
\(=5x^2-3x-x^3+x^2+x^3-6x^2-10x+3x=-10\) => đpcm
25 tháng 1 2021
a) Ta có: \(2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
b) Ta có: \(2x^3+6x^2=x^2+3x\)
\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)
\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)
\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)
\(\Leftrightarrow12x^2+15x-18=0\)
\(\Leftrightarrow12x^2+24x-9x-18=0\)
\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)
H
2 tháng 1 2023
\(a,đk:x\ne0;4;1\)
\(\dfrac{x-1}{x^2-5x+4}-\dfrac{4}{x^2-4x}\\ =\dfrac{x-1}{\left(x-1\right)\left(x-4\right)}-\dfrac{4}{x\left(x-4\right)}\\ =\dfrac{x\left(x-1\right)}{x\left(x-1\right)\left(x-4\right)}-\dfrac{4\left(x-1\right)}{x\left(x-1\right)\left(x-4\right)}\\ =\dfrac{x^2-x-4x+4}{x\left(x-1\right)\left(x-4\right)}\\ =\dfrac{x^2-5x+4}{x.\left(x-1\right)\left(x-4\right)}=\dfrac{\left(x-1\right)\left(x-4\right)}{x.\left(x-1\right)\left(x-4\right)}=\dfrac{1}{x}\)
\(đk:x\ne-2;1\)
\(\dfrac{x}{x+2}+\dfrac{7x-16}{\left(x+2\right)\left(7x-7\right)}\\ =\dfrac{x\left(7x-7\right)}{\left(x+2\right)\left(7x-7\right)}+\dfrac{7x-16}{\left(x+2\right)\left(7x-7\right)}\\ =\dfrac{7x^2-7x+7x-16}{\left(x+2\right)\left(7x-7\right)}\\ =\dfrac{7x^2-16}{\left(x+2\right)\left(7x-7\right)}\)
NH
2 tháng 1 2023
a)
\(\dfrac{x-1}{x^2-5x+4}-\dfrac{4}{x^2-4x}\) \(ĐKXĐ:x\ne0;x\ne4;x\ne1\)
\(=\dfrac{x-1}{x^2-4x-x+4}-\dfrac{4}{x\left(x-4\right)}\)
\(=\dfrac{x-1}{x\left(x-4\right)-\left(x-4\right)}-\dfrac{4}{x\left(x-4\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x-4\right)}-\dfrac{4}{x\left(x-4\right)}\)
\(=\dfrac{x^2-x}{x\left(x-1\right)\left(x-4\right)}-\dfrac{4\left(x-1\right)}{x\left(x-1\right)\left(x-4\right)}\)
\(=\dfrac{x^2-x-4x+4}{x\left(x-1\right)\left(x-4\right)}\)
\(=\dfrac{x\left(x-1\right)-4\left(x-1\right)}{x\left(x-1\right)\left(x-4\right)}\)
\(=\dfrac{\left(x-1\right)\left(x-4\right)}{x\left(x-1\right)\left(x-4\right)}\\ =\dfrac{1}{x}\)
b)
\(\dfrac{x}{x+2}+\dfrac{7x-16}{\left(x+2\right)\left(7x-7\right)}\) \(ĐKXĐ:x\ne-2;x\ne1\)
\(=\dfrac{x\left(7x-7\right)}{\left(x+2\right)\left(7x-7\right)}+\dfrac{7x-16}{\left(x+2\right)\left(7x-7\right)}\)
\(=\dfrac{7x^2-7x+7x-16}{\left(x+2\right)\left(7x-7\right)}\)
\(=\dfrac{7x^2-16}{\left(x+2\right)\left(7x-7\right)}\)
\(a,5x\left(x-1\right)=x-1\)
\(\Rightarrow5x\left(x-1\right)-x+1=0\)
\(\Rightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\5x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}}\)
\(b,2\left(x-7\right)-x^2+7x=0\)
\(\Rightarrow2\left(x-7\right)-x\left(x-7\right)=0\)
\(\Rightarrow\left(x-7\right)\left(2-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-7=0\\2-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=7\\x=2\end{cases}}}\)