tìm x biết
(2x-3)(6-1/5x)=0
Tìm x biết a) (5x - 1) (2x - 1 phần 3)=0 b) 6(x - 1)+ 2x(x - 1)=0
(5x - 1)(2x - 1/3) = 0
<=> 5x - 1 = 0 hoặc 2x - 1/3 = 0
=> x = 1/5 hoặc x = 1/6
vậy x= 1/5 hoặc x= 1/6
Tìm x biết a) (5x - 1) (2x - 1 phần 3)=0 b) 6(x - 1)+ 2x(x - 1)=0
a) (5x - 1) . ( 2x - 1/3 ) = 0
=> 5x - 1 = 0
2x - 1/3 = 0
=> 5x = 1
2x = 1/3
=> x = 1/5
x = 1/6
Tìm x biết a) (5x - 1) (2x - 1 phần 3)=0 b) 6(x - 1)+ 2x(x - 1)=0
a) \(\left(5x-1\right)\left(\frac{2x-1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\2x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{2}\end{cases}}\)
b) \(6\left(x-1\right)+2x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6+2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\6+2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)
a) \(\left(5x-1\right)\cdot\frac{2x-1}{3}=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\\frac{2x-1}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=1\\2x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{2}\end{cases}}}\)
Vậy \(x=\frac{1}{5};x=\frac{1}{2}\)
b) 6(x-1)+2x(x-1)=0
<=> (x-1)(6+2x)=0
<=> \(\orbr{\begin{cases}x-1=0\\6+2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)
Vậy x=1; x=-3
Tìm x,biết
1) 3x^2 - 4x = 0
2) (x^2 - 5x) + x - 5 = 0
3) x^2 - 5x + 6 = 0
4) 5x(x-3) - x+3 = 0
5) x^2 - 2x + 5 = 0
6) x^2 + x -6 = 0
Answer:
\(3x^2-4x=0\)
\(\Rightarrow x\left(3x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)
\(\left(x^2-5x\right)+x-5=0\)
\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)
\(x^2-5x+6=0\)
\(\Rightarrow x^2-2x-3x+6=0\)
\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)
\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
\(5x\left(x-3\right)-x+3=0\)
\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Rightarrow\left(5x-1\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)
\(x^2-2x+5=0\)
\(\Rightarrow\left(x^2-2x+1\right)+4=0\)
\(\Rightarrow\left(x-1\right)^2=-4\) (Vô lý)
Vậy không có giá trị \(x\) thoả mãn
\(x^2+x-6=0\)
\(\Rightarrow x^2+3x-2x-6=0\)
\(\Rightarrow x.\left(x+3\right)-2\left(x+3\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)
I) THỰC HIỆN PHÉP TÍNH a) 2x(x^2-4y) b)3x^2(x+3y) c) -1/2x^2(x-3) d) (x+6)(2x-7)+x e) (x-5)(2x+3)+x II phân tích đa thức thành nhân tử a) 6x^2+3xy b) 8x^2-10xy c) 3x(x-1)-y(1-x) d) x^2-2xy+y^2-64 e) 2x^2+3x-5 f) 16x-5x^2-3 g) x^2-5x-6 IIITÌM X BIẾT a)2x+1=0 b) -3x-5=0 c) -6x+7=0 d)(x+6)(2x+1)=0 e)2x^2+7x+3=0 f) (2x-3)(2x+1)=0 g) 2x(x-5)-x(3+2x)=26 h) 5x(x-1)=x-1 IV TÌM GTNN,GTLN. a) tìm giá trị nhỏ nhất x^2-6x+10 2x^2-6x b) tìm giá trị lớn nhất 4x-x^2-5 4x-x^2+3
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
Tìm x biết
a) 25x^2 -1-(5x-1)(x+2) = 0
b) (2x-3)-(3-2x)(x-1) = 0
c) 9 -4x^2-(6+4x)(x-5) = 0
b) ( 2x - 3 ) - ( 3 - 2x )( x - 1 ) = 0
<=> ( 2x - 3 ) + ( 2x - 3 )( x - 1 ) = 0
<=> ( 2x - 3 )( 1 + x - 1 ) = 0
<=> x( 2x - 3 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}}\)
Vậy .....
a, 25x^2 - 1 - (5x -1)(x+2)=0
=> (5x)^2 - 1 + (5x-1)(x+2) = 0
=> (5x-1)(5x+1) + (5x-1)(x+2) = 0
=> (5x-1)(5x+1+x+2) = 0
=> (5x-1)(6x+3) = 0
=> \(\orbr{\begin{cases}5x-1=0\\6x+3=0\end{cases}}\)
a) 25x2 - 1 - ( 5x - 1 )( x + 2 ) = 0
<=> ( 5x - 1 )( 5x + 1 ) - ( 5x - 1 )( x + 2 ) = 0
<=> ( 5x - 1 )( 5x + 1 - x - 2) = 0
<=> ( 5x - 1 )( 4x - 1 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\4x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{4}\end{cases}}}\)
Vậy .......
Tìm x, biết
a, x (x - 3) - 2x + 6 = 0
b, (3x - 5) (5x - 7) + (5x +1) (2 - 3x)
Ta có : x(x - 3) - 2x + 6 = 0
<=> x(x - 3) - (2x - 6) = 0
=> x(x - 3) - 2(x - 3) = 0
=> (x - 2)(x - 3) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
Tìm x, biết:
6) x^3 - 2x^2 + 2x = 0
7) 2x^3 - 5x^2 + 8x - 5 = 0
Tìm x, biết :
|5x - 4| = |x + 2|
|2x - 3| - |3x + 2| = 0
|5/4. x - 7/2| - | 5/8. x + 3/5| = 0
|7x + 1| - |5x + 6| = 0
|5\(x\) - 4| = |\(x+2\)|
\(\left[{}\begin{matrix}5x-4=x+2\\5x-4=-x-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}4x=6\\6x=2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
vậy \(x\in\) { \(\dfrac{1}{3};\dfrac{3}{2}\)}
|2\(x\) - 3| - |3\(x\) + 2| = 0
|2\(x\) - 3| = | 3\(x\) + 2|
\(\left[{}\begin{matrix}2x-3=3x+2\\2x-3=-3x-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{5}\end{matrix}\right.\)
vậy \(x\in\){ -5; \(\dfrac{1}{5}\)}
|\(\dfrac{5}{4}\)\(x\) - \(\dfrac{7}{2}\)| - | \(\dfrac{5}{8}\)\(x\) + \(\dfrac{3}{5}\)| = 0
|\(\dfrac{5}{4}x\) - \(\dfrac{7}{2}\)| = | \(\dfrac{5}{8}x+\dfrac{3}{5}\)|
\(\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{7}{2}=\dfrac{5}{8}x+\dfrac{3}{5}\\\dfrac{5}{4}x-\dfrac{7}{2}=-\dfrac{5}{8}x-\dfrac{3}{5}\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{5}{8}x=\dfrac{3}{5}+\dfrac{7}{2}\\\dfrac{5}{4}x+\dfrac{5}{2}x=-\dfrac{3}{5}+\dfrac{7}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{5}{8}x=\dfrac{41}{10}\\\dfrac{15}{8}x=\dfrac{29}{10}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{164}{25}\\x=\dfrac{116}{75}\end{matrix}\right.\)
Vậy \(x\in\) { \(\dfrac{116}{75}\); \(\dfrac{164}{25}\)}