tìm x biết a) 2x(x-1)-2x^2=-6 b) 2x(x-3)+5(x-3)=0
c) x^2+x-6=0
a) (2x +1)(3 – x)(4 - 2x) = 0 b)2x(x – 3) + 5(x – 3) = 0
c) (x2 – 4) – (x – 2)(3 – 2x) = 0 d) x2 – 5x + 6 = 0
e) (2x + 5)2 = (x + 2)2 f) 2x3 + 6x2 = x2 + 3x
a: (2x+1)(3-x)(4-2x)=0
=>(2x+1)(x-3)(x-2)=0
hay \(x\in\left\{-\dfrac{1}{2};3;2\right\}\)
b: 2x(x-3)+5(x-3)=0
=>(x-3)(2x+5)=0
=>x=3 hoặc x=-5/2
c: =>(x-2)(x+2)+(x-2)(2x-3)=0
=>(x-2)(x+2+2x-3)=0
=>(x-2)(3x-1)=0
=>x=2 hoặc x=1/3
d: =>(x-2)(x-3)=0
=>x=2 hoặc x=3
e: =>(2x+5+x+2)(2x+5-x-2)=0
=>(3x+7)(x+3)=0
=>x=-7/3 hoặc x=-3
f: \(\Leftrightarrow2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
hay \(x\in\left\{0;-3;\dfrac{1}{2}\right\}\)
Tìm x
a) 3x(4x - 3) - 2x(5 - 6x) = 0
b) 5(2x - 3) + 4x(x - 2) + 2x(3 - 2x) = 0
c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)
d) 3x (x + 1) - 5x(3 - x) + 6(x^2 + 2x + 3) = 0
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
b) 5(2x-3)+4x(x-2)+2x(3-2x)=0
\(\Leftrightarrow\)10x-15+4x2-8x+6x-4x2=0
\(\Leftrightarrow8x-15=0\)
\(\Leftrightarrow8x=15\)
\(\Leftrightarrow x=\dfrac{15}{8}\)
vậy x=\(\dfrac{15}{8}\)
c)3x(2-x)+2x(x-1)=5x(x+3)
\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\\ \Leftrightarrow4x-x^2=5x^2+15x\\ \Leftrightarrow4x-x^2-5x^2-15x=0\\ \)
\(\Leftrightarrow-6x^2-11x=0\\ \Leftrightarrow-x\left(6x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-11}{6}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{-11}{6}\)
1/Tìm x ,y biết
a/(x-1)*(x-2)=-5
2/ tìm x biết
a/2*(x-1)-3*(x-2)=-5-2*(x-3)
b/x*(x-2) < 0c
c/(x-3)*(2x+6)*(3x-15)=0
1/a)Ta có -5=-1*5=-5*1
Sau đó bn lập bảng rồi dùng ước bội để tính nhé
Tk mình nha bn!
a) (2x+1)^2-2x-1=2b) (x^2-3x)^2+5(x^2-3x)+6=0c) (x^2-x-1)(x^2-x)-2=0d) (5-2x)^2+4x-10=8e) (x^2+2x+3)(x^2+2x+1)=3f) x(x-1)(x^2-x+1)-6=0
sửa lại chút: a) (2x+1)^2-2x-1=2 b) (x^2-3x)^2+5(x^2-3x)+6=0 c) (x^2-x-1)(x^2-x)-2=0 d) (5-2x)^2+4x-10=8 e) (x^2+2x+3)(x^2+2x+1)=3 f) x(x-1)(x^2-x+1)-6=0
a) Ta có: \(\left(2x+1\right)^2-2x-1=2\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x+1\right)+\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1-2\right)+\left(2x+1-2\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{1}{2};-1\right\}\)
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:
a) 4.(2-x)+x.(x+6)=x2
b) x.(x-7)-(x-2).(x+5)=0
c) (2x+3).(3-2x)+(2x-1)2=2
a: Ta có: \(4\left(2-x\right)+x\left(x+6\right)=x^2\)
\(\Leftrightarrow8-4x+x^2+6x-x^2=0\)
\(\Leftrightarrow2x=-8\)
hay x=-4
b: Ta có: \(x\left(x-7\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow x^2-7x-x^2-3x+10=0\)
\(\Leftrightarrow-10x=-10\)
hay x=1
c: Ta có: \(\left(2x+3\right)\left(3-2x\right)+\left(2x-1\right)^2=2\)
\(\Leftrightarrow9-4x^2+4x^2-4x+1=2\)
\(\Leftrightarrow-4x=-8\)
hay x=2
Bài 5: Tìm x (Giải phương trinh)
a)x^3-13x=0
b) 5x(x – 2000) – x + 2000 = 0
c) 2x(x – 2) + 3(x – 2) = 0
d) x + 1 = (x + 1)2
e) x + 5x2 = 0
f) x3 + x = 0
Bài 5: Tìm x (Giải phương trình)
a)x^3-13x=0 b) 5x(x – 2000) – x + 2000 = 0
c) 2x(x – 2) + 3(x – 2) = 0 d) x + 5x2 = 0
d) x + 1 = (x + 1)2 e) x3 + x = 0
b) 5x(x-2000)-x+2000=0
\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\\ \Rightarrow\left(x-2000\right)\left(5x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-2000=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+2000\\5x=0+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\5x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\)
c) Ta có: \(2x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-3}{2}\end{matrix}\right.\)
d) Ta có: \(5x^2+x=0\)
\(\Leftrightarrow x\left(5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{5}\end{matrix}\right.\)
Giải các phương trình sau: a. 2x-5/x+5=3
b. (x^2+2x)-(3x+6)/x-3=0
c. x/2(x-3)+x/2x+2=2x/(x+1)(x-3)
a, ĐKXĐ:\(x\ne-5\)
\(\dfrac{2x-5}{x+5}=3\\ \Rightarrow2x-5=3\left(x+5\right)\\ \Leftrightarrow3x+15-2x+5=0\\ \Leftrightarrow x+20=0\\ \Leftrightarrow x=-20\)
b, ĐKXĐ:\(x\ne3\)
\(\dfrac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\\ \Rightarrow x^2+2x-3x-6=0\\ \Leftrightarrow x^2-x-6=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(3x+6\right)=0\\ \Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)
c, ĐKXĐ:\(\left\{{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)
\(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\\ \Leftrightarrow x\left(\dfrac{1}{2\left(x-3\right)}+\dfrac{1}{2\left(x+1\right)}-\dfrac{2}{\left(x+1\right)\left(x-3\right)}\right)=0\\ \Leftrightarrow x\left(\dfrac{x+1}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x-3}{2\left(x+1\right)\left(x-3\right)}-\dfrac{4}{2\left(x+1\right)\left(x-3\right)}\right)=0\\ \Leftrightarrow x.\dfrac{x+1+x-3-4}{2\left(x-3\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x\left(2x-6\right)}{2\left(x-3\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{2x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x}{x+1}=0\\ \Rightarrow x=0\left(tm\right)\)
Tìm x biết: (x + 2)^2 - (x + 2)(x - 3) = 0
Tìm x biết :
a,(x+2)^2-(x+2)(x-3)=0
b,2x^3-4x^2+2x=0
c,(x-1)^2-(2x+1)^2=0
\(a,\Leftrightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Leftrightarrow5\left(x+2\right)=0\Leftrightarrow x=-2\\ b,\Leftrightarrow2x\left(x-1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ c,\Leftrightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Leftrightarrow3x\left(-x-2\right)=0\Leftrightarrow-3x\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)