2x*(48-12x)-10*(48-12x)=0
2x*(48-12x)-10*(48-12x)=0
2x*(48-12x)-10*(48-12x)=0
=> (48-12x).(2x-10)=0
=> 48-12x=0 hoặc 2x-10=0
=> 12x=48-0 hoặc 2x =0+10
=> 12x=48 hoặc 2x =10
=> x=48:12 hoặc x =10:2
=> x=4 hoặc x =5
1, tìm x biết
a, (2x-1)^2-2x+1=0
b, (4x-3)^2-12x+9=0
c, 3x^2-48=0
a) \(\left(2x-1\right)^2-2x+1=0\)
\(\Leftrightarrow4x^2-4x+1-2x+1=0\)
\(\Leftrightarrow4x^2-6x+2=0\)
\(\Delta=\left(-6\right)^2-4.4.2=4\)
\(x_1=\dfrac{1}{2};x_2=1\)
b) \(\left(4x-3\right)^2-12x+9=0\)
\(\Leftrightarrow16x^2-24x+9-12x+9=0\)
\(\Leftrightarrow16x^2-36x+18=0\)
\(\Delta=\left(-36\right)^2-4.16.18=144\)
\(x_1=\dfrac{3}{4};x_2=\dfrac{3}{2}\)
c) \(3x^2-48=0\)
\(\Leftrightarrow3x^2=48\)
\(\Leftrightarrow x^2=16\)
\(x_1=4;x_2=-4\)
Tick cho mk nha
Tìm x:
a)x^3+1=0
b)6x^2-12x-48=0
a: Ta có: \(x^3+1=0\)
\(\Leftrightarrow x^3=-1\)
hay x=-1
b: Ta có: \(6x^2-12x-48=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
Giải phương trình sau
a) \(\sqrt{\left(2x-5\right)^2}=7\)
b) \(\sqrt{3x}-\sqrt{12x}=\sqrt{27}-\sqrt{48}\)
(a) Phương trình tương đương: \(\left|2x-5\right|=7\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=7\\2x-5=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\).
Vậy: \(S=\left\{-1;6\right\}\)
(b) Điều kiện: \(x\ge0\).
Phương trình tương đương: \(\sqrt{3x}-2\sqrt{3x}=3\sqrt{3}-4\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}\left(\sqrt{x}-2\sqrt{x}\right)=-\sqrt{3}\)
\(\Leftrightarrow\sqrt{x}-2\sqrt{x}=-1\)
\(\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\left(TM\right)\).
Vậy: \(S=\left\{1\right\}\)
12x- 15y/7= 20z-12x/9=15y-20z/11 và x+y+z= 48
\(\frac{12x-5y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}=\frac{12x-15y+20z-12x+15y-20x}{7+9+11}=0\)\(=0\)
=> 12x - 15y =0
=> 12x = 15y
=> \(\frac{x}{15}=\frac{y}{12}\)
=> \(\frac{x}{60}=\frac{y}{48}\)
20z - 12x = 0
=> 20z = 12x
=> \(\frac{x}{20}=\frac{z}{12}\)
=> \(\frac{x}{60}=\frac{z}{36}\)
=> \(\frac{x}{60}=\frac{y}{48}=\frac{z}{36}=\frac{x+y+z}{60+48+36}=\frac{48}{144}=\frac{1}{3}\)
=> x = 1 . 60 : 3 = 20
y = 1 . 48 : 3 = 16
z = 1 . 36 : 3 = 12
12x-15y/7=20z-12x/9=15y-20x/11 và x+y+z=48
tìm x,y,z biết: 12x-15y/7= 20z-12x/9= 15y-20z/11 và x+y+z=48
(12x-48)^2=144
(12x - 48)^2 = 114
(12x - 48)^2 = 12^2
(12x - 48)= 12
12x = 12+48
12x = 60
x = 60/12
x = 5
(12x-48)2=144
(12x-48)2= 122 ( cũng có thể nói 12x-48 = \(\sqrt{144}\))
=>12x-48=12
12x = 12 + 48 = 60
x = 60:12
x = 5
tìm x, ý, z biết 12x-15y/7=20z-12x/9=15y-20z/11 và x+y+z= 48