tìm x : I 5x - 1 I - I x - 4 I = 0
tìm x biết :
a, I 5x -4 I = I x+2 I
b, I x + \(\frac{2}{5}\) I = 2x
c, I x -1 I + I x - 3 I = 2x - 1
d, I 5x +1 I + I 6y -8 I < hoặc bằng 0
Bài làm:
a) | 5x - 4 | = | x + 2 |
=> 5x - 4 = x + 2
=> 5x - x = 2 + 4
=> x . (5 - 1) = 6
=> x . 4 = 6
=> x = 6 : 4 = 1,5
b) | x + 2/5 | = 2x
=> x + 2/5 = 2x hoặc x + 2/5 = -2x
* x + 2/5 = 2x
=> x - 2x = -2/5
=> x . (1 - 2) = -2/5
=> x .(-1) = -2/5
=> x = -2/5 : (-1)
=> x = 2/5
* x + 2/5 = -2x
=> x + 2x = 2/5
=> x . (1 + 2) = 2/5
=> x . 3 = 2/5
=> x = 2/5 : 3
=> x = 2/15
mk chỉ làm 2 bài này thôi, còn 2 bài kia mk ko có pít làm. Sorry!
tìm x :
xy-10x-12y+128 =0
xy-2x+4y=10
I x +1 I + I x+2 I + I x+3 I + I x+4 I =5x+5
ai nhanh thì mk sẽ tk ! nhanh giùm mk nha! mk mai phải nộp rồi !
I 7+5x I = 1-4x
I 4x^2 - 2x I + 1 = 2x
I x^2 - 5x + 4 I = x+4
I 4 - 3x I = 3x -4
I 1+5x I = 1 + 5x
I x^2 - 3x + 1 I = 2x-3
I x-1 I = x^2 -x
|7 + 5x| = 1 - 4x
=> \(\orbr{\begin{cases}7+5x=1-4x\left(đk:x\le\frac{1}{4}\right)\\7+5x=4x-1\left(đk:x\ge\frac{1}{4}\right)\end{cases}}\)
=> \(\orbr{\begin{cases}7-1=-4x-5x\\7+1=4x-5x\end{cases}}\)
=> \(\orbr{\begin{cases}6=-9x\\8=-x\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{2}{3}\left(tm\right)\\x=-8\left(ktm\right)\end{cases}}\)
|4x2 - 2x| + 1 = 2x
=> |4x2 - 2x| = 2x - 1
=> \(\orbr{\begin{cases}4x^2-2x=2x-1\left(đk:x\ge\frac{1}{2}\right)\\4x^2-2x=1-2x\left(đk:x\le\frac{1}{2}\right)\end{cases}}\)
=> \(\orbr{\begin{cases}4x^2-2x-2x+1=0\\4x^2-2x-1+2x=0\end{cases}}\)
=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\4x^2-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x-1=0\\x^2=\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=\pm\frac{1}{2}\end{cases}}\)(tm)
Vậy ...
1. Tìm x
a). I x I + I x + 1 I + I x + 2 I + I x + 3 I + I x + 4 I = 5x
b). ( 2x - 5 ) - ( 3x - 7 ) = x+ 3
b) Theo bài ra , ta có :
(2x - 5) - (3x - 7) = x + 3
(=) 2x - 5 - 3x + 7 = x + 3
(=) -2x = 1
(=) x = -1/2
Vậy x = -1/2
Chúc bạn học tốt =))
Giải các bất phương trình sau A, -x2 + 5x - 4 ≤ 0
B, x2+5x+4 >0
\(-x^2+5x-4\le0\)
\(\Leftrightarrow\left(x-1\right)\left(x-4\right)\ge0\Rightarrow\left[{}\begin{matrix}x\ge4\\x\le1\end{matrix}\right.\)
\(x^2+5x+4>0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)>0\Rightarrow\left[{}\begin{matrix}x>-1\\x< -4\end{matrix}\right.\)
tìm x : I x + 3 I+ 10 - 5x = 0
`|x+3|+10-5x=0`
`<=>|x+3|=5x-10(x>=2)`
`+)x+3=5x-10`
`<=>4x=13`
`<=>x=13/4(tm)`
`+)x-3=10-5x`
`<=>6x=13`
`<=>x=13/6(tm)`
Vậy `S={13/4,13/6}`
\(\left|x+3\right|+10-5x=0\)
\(\Leftrightarrow\left|x+3\right|=5x-10\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x-10\ge0\\\left[{}\begin{matrix}x+3=5x-10\\x+3=10-5x\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left[{}\begin{matrix}x=\dfrac{13}{4}\left(N\right)\\x=\dfrac{7}{6}\left(L\right)\end{matrix}\right.\end{matrix}\right.\)
Giải:
\(\left|x+3\right|+10-5x=0\)
\(\Rightarrow\left|x+3\right|=5x-10\)
\(\Rightarrow\left[{}\begin{matrix}5x-10=x+3\\5x-10=x-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{4}\\x=\dfrac{7}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{13}{4};\dfrac{7}{4}\right\}\)
Chúc bạn học tốt!
tìm x : I x + 3 I+ 10 - 5x = 0
`|x+3|+10-5x=0`
`<=>|x+3|=5x-10(x>=2)`
`+)x+3=5x-10`
`<=>4x=13`
`<=>x=13/4(tm)`
`+)x-3=10-5x`
`<=>6x=13`
`<=>x=13/6(tm)`
Vậy `S={13/4,13/6}`
TH1: `x+3>=0 <=> x>=-3`
`x+3+10-5x=0`
`-4x=-13`
`x=13/4` (TM)
TH2: `x+3<0 <=> x<-3`
`-x-3+10-5x=0`
`-6x=-7`
`x=7/6` (L)
Vậy `x=13/4`
a) (x+1/x-2)^2 + x+1/x-4 -3(2x-4/x-4)^2 = 0
b) 15x/x^2 +3x-4 - 1 = 12(1/x+4 + 1/3x-3)
c) x^2-4x+1/x+1 + 2 = - x^2-5x+1/2x+1
Tìm x,biết
I 2x^2+4x I+I x^2+5x+6 I=0
\(\left|2x^2+4x\right|+\left|x^2+5x+6\right|=0.^{\left(1\right)}\)
\(NX\hept{\begin{cases}\left|2x^2+4x\right|\ge0\\\left|x^2+5x+6\right|\ge0\end{cases}\Rightarrow}\left(1\right)\ge0\)
Dấu \("="\)xảy ra khi và chỉ khi
\(\hept{\begin{cases}\left|2x^2+4x\right|=0\\\left|x^2+5x+6\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x^2+4x=0\\x^2+5x+6=0\end{cases}\Leftrightarrow\hept{\begin{cases}x\left(2x+4\right)=0\\x\left(x+5\right)=0-6\end{cases}}}\Leftrightarrow\hept{\begin{cases}x=0;x=-2\\x\inƯ\left(6\right)\end{cases}\Rightarrow x=-2}\)
Vậy x = -2
\(\left|2x^2+4x\right|+\left|x^2+5x+6\right|=0\)
Ta có : \(\hept{\begin{cases}\left|2x^2+4x\right|\ge0\\\left|x^2+5x+6\right|\ge0\end{cases}}\Rightarrow\left|2x^2+4x\right|+\left|x^2+5x+6\right|\ge0\)
\(\Rightarrow\orbr{\begin{cases}2x^2+4x=0\\x^2+5x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\left(2x+4\right)=0\left(1\right)\\x\left(x+5\right)=-6\left(2\right)\end{cases}}\)
(1) \(x\left(2x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
(2) x(x+5)=-6
=> x2+5x=-6
=> x2+5x+6=0
=> x2 +3x+2x+6=0
=> x(x+3)+2(x+3) = 0
=> (x+3)(x+2)=0
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=-2\end{cases}}\)
Vậy ........