Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left|x-5\right|=2x\)ĐK : x>=0
TH1 : x - 5 = 2x <=> x = -5 ( loại )
TH2 : x - 5 = -2x <=> 3x = 5 <=> x = 5/3 ( tm )
Vậy tập nghiệm pt là S = { 5/3 }
\(\left(x-2\right)^2+2\left(x-1\right)\le x^2+4\)
\(\Leftrightarrow x^2-4x+4+2x-2-x^2-4\le0\)
\(\Leftrightarrow-2x-2\le0\Leftrightarrow x+1\ge0\Leftrightarrow x\ge-1\)
Vậy tập nghiệm bft là S = { x | x > = -1 }
Ta có: \(\left|x-5\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=2x\left(x\ge5\right)\\x-5=-2x\left(x< 5\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2x=5\\x+2x=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=5\\3x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\left(loại\right)\\x=\dfrac{5}{3}\left(nhận\right)\end{matrix}\right.\)
`(-7x^2+4)/(x^3+1)=5/(x^2-x+1)-1/(x+1)(x ne -1)`
`<=>-7x^2+4=5(x+1)-x^2+x-1`
`<=>-7x^2+4=5x+5-x^2+x-1`
`<=>6x^2+6x=0`
`<=>6x(x+1)=0`
Vì `x ne -1=>x+1 ne 0`
`=>x=0`
Vậy `S={0}`
ĐKXĐ: \(x\ne-1\)
Ta có: \(\dfrac{-7x^2+4}{x^3+1}=\dfrac{5}{x^2-x+1}-\dfrac{1}{x+1}\)
\(\Leftrightarrow\dfrac{5\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{-7x^2+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
Suy ra: \(5x+5-x^2+x-1=-7x^2+4\)
\(\Leftrightarrow-x^2+6x+4+7x^2-4=0\)
\(\Leftrightarrow6x^2+6x=0\)
\(\Leftrightarrow6x\left(x+1\right)=0\)
mà 6>0
nên x(x+1)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)
Vậy: S={0}
ĐKXĐ: ` x ne 1 ; x ne 4`
`(2x+1)/(x^2-5x+4) + 5/(x-1) = 2/(x-4)`
`<=> (2x+1)/[(x-1)(x-4)] + [5(x-4)]/[(x-1)(x-4)] = [2(x-1)]/[(x-1)(x-4)]`
`=> 2x+1 + 5x -20 = 2x-2`
`<=> 5x = 17`
`<=> x= 17/5`(thỏa mãn ĐKXĐ)
Vậy tập nghiệm của phương trình là `S={ 17/5}`
Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
ĐKXĐ: \(x\ne\pm2\)
\(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x+2\right)\left(x-2\right)}+\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\\ \Leftrightarrow\left(x+1\right)\left(x+2\right)-5\left(x-2\right)=12+\left(x+2\right)\left(x-2\right)\\ \Leftrightarrow x^2+x+2x+2-5x+10=12+x^2-4\\ \Leftrightarrow-2x=-4\\ \Leftrightarrow x=2\left(ktm\right)\)
Vậy \(S\in\left\{\varnothing\right\}\)
ĐKXĐ: \(\begin{cases}x-2\ne 0\\x+2\ne 0\end{cases}\leftrightarrow x\ne 2\\x\ne -2\end{cases}\)
\(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\leftrightarrow \dfrac{(x+1)(x+2)}{(x-2)(x+2)}-\dfrac{5(x-2)}{(x+2)(x-2)}=\dfrac{12}{(x-2)(x+2)}+\dfrac{(x-2)(x+2)}{(x-2)(x+2)}\)
\(\to x^2+3x+2-5x+10=12+x^2-4\)
\(\leftrightarrow x^2-2x-x^2=12-12-4\)
\(\leftrightarrow -2x=-4\)
\(\leftrightarrow x=2(\rm KTM)\)
Vậy pt đã cho vô nghiệm \(S=\varnothing\)
`5-(x-6)=4(3-2x)`
`<=>5-x+6-4(3-2x)=0`
`<=> 5-x+6-12 +8x=0`
`<=> 7x -1=0`
`<=> 7x=1`
`<=>x=1/7`
Vậy pt đã cho có nghiệm `x=1/7`
__
`3-x(1-3x) =5(1-2x)`
`<=> 3-x+3x^2=5-10x`
`<=> 3-x+3x^2-5+10x=0`
`<=> 3x^2 +9x-2=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-9+\sqrt{105}}{6}\\x=\dfrac{-9-\sqrt{105}}{6}\end{matrix}\right.\)
Vậy pt đã cho có tập nghiệm \(S=\left\{\dfrac{-9+\sqrt{105}}{6};\dfrac{-9-\sqrt{106}}{5}\right\}\)
__
`(x-3)(x+4) -2(3x-2)=(x-4)^2`
`<=>x^2+4x-3x-12- 6x +4 =x^2 -8x+16`
`<=>x^2-5x-8=x^2-8x+16`
`<=> x^2 -5x-8-x^2+8x-16=0`
`<=> 3x-24=0`
`<=>3x=24`
`<=>x=8`
Vậy pt đã cho có nghiệm `x=8`
a) 5-(x-6)=4(3-2x)
=> 5 – x + 6 = 12 – 8x
=> -x + 8x = 12 – 5 – 6
=> 7x = 1
=> x=1/7
Vậy phương trình có nghiệm x=1/7
b) 3 - x ( 1 - 3x)=5(1-2x)
=> 3-x+3x^2=5-10x
=> 3x^2+9x-2= 0
0=105
=> x =\(\dfrac{-9-\sqrt{105}}{6}\)
1/
-x^3 -5x^2 + 4x +4
=> x1 =-5.5877............
x2=1.1895.............
x3=-0.6018............
a: =>4x^2-24x+36-4x^2+4x-1<10
=>-20x<10-35=-25
=>x>=5/4
b: =>x(x^2-25)-x^3-8<=3
=>x^3-25x-x^3-8<=3
=>-25x<=11
=>x>=-11/25
\(\frac{x+1}{x+2}+\frac{5}{x-2}=\frac{4}{x-4+1}\left(ĐKXĐ:x\ne\pm2,x\ne3\right)\)
<=> \(\frac{x+1}{x+2}+\frac{5}{x-2}=\frac{4}{x-3}\)
<=> \(\frac{\left(x+1\right)\left(x-3\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)\left(x-3\right)}+\frac{5\left(x-3\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)\left(x-3\right)}=\frac{4\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)\left(x-3\right)}\)
<=> (x + 1)(x - 3)(x - 2) + 5(x - 3)(x + 2) = 4(x - 2)(x + 2)
<=> (x2 - 3x + x - 3)(x - 2) + 5(x2 + 2x - 3x - 6) = 4(x2 - 4)
<=> x3 - 2x2 - 3x2 + 6x + x2 - 2x - 3x + 6 + 5x2 + 10x - 15x - 30 = 4x2 - 16
<=> x3 - 2x2 - 3x2 + 6x + x2 - 2x - 3x + 6 + 5x2 + 10x - 15x - 30 - 4x2 + 16 = 0
<=> x3 - 3x2 - 4x - 8 = 0
PT vô nghiệm
Vậy \(S=\varnothing\)