tim x biet : 2|4x-2| - 2x = 14
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9 tháng 8 2015
a) x^2 - 11x + 18 = 0
=> x^2 - 2x - 9x + 18 = 0
=> x ( x- 2 ) - 9 ( x- 2 ) = 0
=> ( x- 9 )( x- 2 )= 0
=> x- 9 = 0 hoặc x - 2 = 0
=> x= 9 hoặc x = 2
S
7 tháng 10 2018
\(a.\sqrt{\left(2x-5\right)^2}=7\Leftrightarrow2x-5=7\)
\(\Leftrightarrow2x=12\Leftrightarrow x=6\)
b, Máy mình lỗi font nên không làm đc
NH
29 tháng 12 2019
Ta có: \(4x=3y\)\(\Rightarrow\frac{x}{3}=\frac{y}{4}\)\(\Rightarrow\frac{x}{15}=\frac{y}{20}\)
\(7y=5z\)\(\Rightarrow\frac{y}{5}=\frac{z}{7}\)\(\Rightarrow\frac{y}{20}=\frac{z}{28}\)
\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=k\)\(\Rightarrow\hept{\begin{cases}x=15k\\y=20k\\z=28k\end{cases}}\)
Ta có: \(yz-2x^2=110\)
\(\Rightarrow20k.28k-2.\left(15k\right)^2=110\)
\(\Rightarrow560k^2-2.225k^2=110\)
\(\Rightarrow560k^2-450k^2=110\)
\(\Rightarrow k^2\left(560-450\right)=110\)
\(\Rightarrow110k^2=110\)
\(\Rightarrow k^2=1\)
\(\Rightarrow\orbr{\begin{cases}k=1\\k=-1\end{cases}}\)
+) Khi k = 1, ta có: \(\hept{\begin{cases}x=15k\\y=20k\\z=28k\end{cases}}\Rightarrow\hept{\begin{cases}x=15.1\\y=20.1\\z=28.1\end{cases}}\Rightarrow\hept{\begin{cases}x=15\\y=20\\z=28\end{cases}}\)
+) Khi k = -1, ta có: \(\Rightarrow\hept{\begin{cases}x=15k\\y=20k\\z=28k\end{cases}}\Rightarrow\hept{\begin{cases}x=15.\left(-1\right)\\y=20.\left(-1\right)\\z=28.\left(-1\right)\end{cases}}\Rightarrow\hept{\begin{cases}x=-15\\y=-20\\z=-28\end{cases}}\)
Vậy...
29 tháng 12 2019
Ta có: \(4x=3y\rightarrow\frac{x}{3}=\frac{y}{4}\rightarrow\frac{x}{15}=\frac{y}{20}\left(1\right)\)
\(7y=5z\rightarrow\frac{y}{5}=\frac{z}{7}\rightarrow\frac{y}{20}=\frac{z}{28}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)
Đặt \(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=k\left(k\varepsilonℕ^∗\right)\)
=> x = 15k; y = 20k; z = 28k
Có: \(yz-2x^2=110\)
\(\Rightarrow20k\cdot28k-2\cdot(15k)^2=110\)
\(\Rightarrow560\cdot k^2-2\cdot225\cdot k^2=110\)
\(\Rightarrow560\cdot k^2-450\cdot k^2=110\)
\(\Rightarrow\left(560-450\right)\cdot k^2=110\)
\(\Rightarrow110\cdot k^2=110\) \(\Rightarrow k^2=1\)
\(\Rightarrow\orbr{\begin{cases}k=1\\k=-1\end{cases}}\)
\(x=15k\rightarrow\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)
\(y=20k\rightarrow\orbr{\begin{cases}y=20\\y=-20\end{cases}}\)
\(z=28k\rightarrow\orbr{\begin{cases}z=28\\z=-28\end{cases}}\)
Vậy...........................
NX
Tim x,
a,2x^4-6x^3+x^2+6x-3=0
b,x^3-9x^2+26x+24=0
c, P= 2x^4 - 4x^3 + 6x^2 - 4x + 5 biet rang x^2 - x=7
1
28 tháng 11 2016
a)\(2x^4-6x^3+x^2+6x-3=0\)
\(\Leftrightarrow2x^4-6x^3+3x^2-2x^2+6x-3=0\)
\(\Leftrightarrow x^2\left(2x^2-6x+3\right)-\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(2x^2-6x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+1=0\\2x^2-6x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\\Delta_{2x^2-6x+3}=\left(-6\right)^2-4\left(2.3\right)=12\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\x_{1,2}=\frac{6\pm\sqrt{12}}{4}\end{array}\right.\)
b)\(x^3+9x^2+26x+24=0\)
\(\Leftrightarrow x^3+5x^2+6x+4x^2+20x+24=0\)
\(\Leftrightarrow x\left(x^2+5x+6\right)+4\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x^2+5x+6\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x+3=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=-3\\x=-4\end{array}\right.\)
5 tháng 2 2018
1.a)x+7=-5-14
x+7=-19
x=-19-7
x=-26
Vậy x=-26
b)3x-4=(-2)3-11
3x-8=-11
3x=-11+8
3x=-3
x=-3:3
X=-1
Vậy x=-1
c)11+(4x-11)=-9-(-15)
11+(4x-11)=-9+15
11+(4x-11)=6
4x-11=6-11
4x=6
x=\(\dfrac{6}{4}\)
Vậy x=\(\dfrac{6}{4}\)
d)\(\left|2x-1\right|=5\)
⇒2x-1=\(\pm5\)
+Nếu 2x-1=5
2x=6
x=6:2
x=3
Vậy x=3
+Nếu 2x-1=-5
2x=-5+1
2x=-4
x=-4:2
x=-2
Vậy xϵ{3;-2}
e)\(\left|x-7\right|+3=25\)
\(\left|x-7\right|=22\)
⇒x-7=\(\pm22\)
+Nếu x-7=22
x=22+7=29
+Nếu x-7=-22
x=-22+7
x=-15
Vậy xϵ{29;-15}
\(\left[{}\begin{matrix}2\left(4x-2\right)-2x=14\left(x\ge\dfrac{1}{2}\right)\\-2\left(4x-2\right)-2x=14\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}6x=18\left(x\ge\dfrac{1}{2}\right)\\-10x=10\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(x\ge\dfrac{1}{2}\right)\left(TM\right)\\x=-1\left(x< \dfrac{1}{2}\right)\left(TM\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)