Bài1: Tìm x biết:
a) x^2-6x+9=4. b) x^2(x-3)-(4X-12)=0
c) (2x+3)^2-4(x+2)^2=12
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(B=\dfrac{3x\left(2x-3\right)-4\left(2x+3\right)-4x^2+23x+12}{\left(2x-3\right)\left(2x+3\right)}\cdot\dfrac{2x+3}{x+3}\)
\(=\dfrac{6x^2-9x-8x-12-4x^2+23x+12}{2x-3}\cdot\dfrac{1}{x+3}\)
\(=\dfrac{2x^2+6x}{\left(2x-3\right)}\cdot\dfrac{1}{x+3}=\dfrac{2x}{2x-3}\)
b: 2x^2+7x+3=0
=>(2x+3)(x+2)=0
=>x=-3/2(loại) hoặc x=-2(nhận)
Khi x=-2 thì \(A=\dfrac{2\cdot\left(-2\right)}{-2-3}=\dfrac{-4}{-7}=\dfrac{4}{7}\)
d: |B|<1
=>B>-1 và B<1
=>B+1>0 và B-1<0
=>\(\left\{{}\begin{matrix}\dfrac{2x+2x-3}{2x-3}>0\\\dfrac{2x-2x+3}{2x-3}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3< 0\\\dfrac{4x-3}{2x-3}>0\end{matrix}\right.\Leftrightarrow x< \dfrac{3}{4}\)
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
f: Ta có: \(x^3-6x^2+12x-19=0\)
\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)
\(\Leftrightarrow\left(x-2\right)^3=11\)
hay \(x=\sqrt[3]{11}+2\)
a)
<=> 3x - 3 + x - 2 = 2x - 2 - x + 1
<=> 3x + x - 2x + x = -2 + 1 + 3 + 2
<=> 3x = 4
<=> x = 4/3
Các câu sau làm tương tự
\(\left(3x-3\right)+\left(x-2\right)=\left(2x-2\right)-\left(x-1\right)\)
<=> \(3x-3+x-2=2x-2-x+1\)
<=> \(4x-5=x-1\)
<=> \(3x=4\)
<=> \(x=\frac{4}{3}\)
Vậy....
a, 4x+1=13-2x <-->6x=12 <-->x=2
b, (2x-5)(x-4)=0 <-->x=5/2 hoặc x=4
c,Đề bài -->x(x-2)+6(x+2)=2x+12 -->x^2+2x=0 -->x=0 hoặc x=-2
d,|x-3|=9-2x -->TH1: x-3=9-2x -->x=x=4 TH2:3-x=9-2x -->x=6
a) `sqrt(x^2-6x _9) = 4-x`
`<=> sqrt[(x-3)^2] =4-x`
`<=> |x-3| =4-x ( đk :x<=4)`
`<=> |x-3| = |4-x|`
`<=> [(x-3 =4-x),(x-3 = x-4):}`
`<=>[(x = 7/2(t//m)),(0=-1(vl)):}`
Vậy `S = {7/2}`
b) `sqrt(x^2 -9) + sqrt(x^2 -6x +9) =0(đk : x>=3(hoặc) x<=-3)`
`<=>sqrt(x^2 -9) =- sqrt(x^2 -6x +9) `
`<=>(sqrt(x^2 -9))^2 =(- sqrt(x^2 -6x +9))^2`
`<=> x^2 -9 = x^2 -6x +9`
`<=> 6x = 9+9 =18`
`<=> x=3(t//m)`
Vậy `S={3}`
c) `sqrt(x^2 -2x+1) + sqrt(x^2-4x+4) =3`
`<=> sqrt[(x-1)^2] +sqrt[(x-2)^2] =3`
`<=> |x-1| +|x-2| =3`
xét `x<1 =>{(|x-1| =1-x ),(|x-2|=2-x):}`
`=> 1-x +2-x =3`
`=> x = 0(t//m)`
xét `1<=x<2 => {(|x-1|=x-1),(|x-2|= 2-x):}`
`=> x-1 +2-x =3`
`=>1=3 (vl)`
xét `x>=2 => {(|x-1| =x-1),(|x-2|=x-2):}`
`=> x-1+x-2 =3`
`=> x=3(t//m)`
Vậy `S = {0;3}`
2:
a: =>x-1=0 hoặc 3x+1=0
=>x=1 hoặc x=-1/3
b: =>x-5=0 hoặc 7-x=0
=>x=5 hoặc x=7
c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)
d: =>x=0 hoặc x^2-1=0
=>\(x\in\left\{0;1;-1\right\}\)
a: \(A=\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{2}{x-2}\)
b: \(B=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}=\dfrac{x+3}{x-3}\)
c: \(C=\dfrac{\left(3x-4\right)\left(3x+4\right)}{x\left(3x-4\right)}=\dfrac{3x+4}{x}\)
d: \(D=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}=\dfrac{x+2}{2}\)
e: \(E=\dfrac{-x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{-x}{x+2}\)
f: \(F=\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}\)
a, \(x^2-6x+9=4< =>\left(x-3\right)^2=4< =>\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)
\(< =>\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
b,\(x^2\left(x-3\right)-4\left(x-3\right)=0< =>\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)
\(< =>\orbr{\begin{cases}x=2\\x=-2\end{cases}orx=3}\)
c nhường mấy bn khácccc
a) x^2-6x+9=4.
x=1, x=5
b) x^2(x-3)-(4X-12)=0
x=-2, x=2, x=3
c) (2x+3)^2-4(x+2)^2=12
x=-19/4