Tìm x biết:
a) x 2 + 3 x = 0 b) x ( 2x − 1) + 4x − 2=0 c) ( x 2 + 2 x )2 − 2 x 2 − 4 x = 3
tìm x: part 1 : a,(x^3)^2-(x+1)(x-1)=1 b,(x-2)^2-3(x-2)=0 c,(x+2)(x^2-2x+4)-x(x^2+2)=15 d,(x+1)^2-(x+1)(x-2)=0 e,4x(x-2017)-x+2017=0 f,(x+4)^2-16=0 part 2: a,x^3+27+(x+3)(x-9)=0 b,(2x-1)^2-4x^2+1=0 c,2(x-3)+x^2-3x=0 d,x^2-2x+1=6x-6 e,x^3-9x=0
tìm x biết
a/ x^3-x^2-x+1=0
b/(2x^3-3)^2-(4x^2-9)=0
c/x^4+2x^3-6x-9=0
d/2(x+5)-x^2-5x=0
\(a)\)\(x^3-x^2-x+1=0\)
\(\Leftrightarrow\)\(x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)^2\left(x+1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(x-1\right)^2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}}\)
Vậy \(x=1\) hoặc \(x=-1\)
Chúc bạn học tốt ~
a) x3-x2-x+1 = 0 \(\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\)\(\Leftrightarrow x^2-1=0\)hoặc x-1=0
\(\Leftrightarrow x=1\)
\(c)\)\(x^4+2x^3-6x-9=0\)
\(\Leftrightarrow\)\(\left(x^4-9\right)+\left(2x^3-6x\right)=0\)
\(\Leftrightarrow\)\(\left(x^2-3\right)\left(x^2+3\right)+2x\left(x^2-3\right)=0\)
\(\Leftrightarrow\)\(\left(x^2-3\right)\left(x^2+3+2x\right)=0\)
\(\Leftrightarrow\)\(x^2-3=0\)
Hoặc \(x^2+3+2x=0\)
\(\Leftrightarrow\)\(x^2=3\)
Hoặc \(x\left(x+2\right)=-3\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
Hoặc \(x;\left(x-2\right)\inƯ\left(-3\right)\)
Ta có bảng :
\(x\) | \(1\) | \(-3\) | \(-1\) | \(3\) |
\(x-2\) | \(-3\) | \(1\) | \(3\) | \(-1\) |
\(x\) | \(1\) | \(-3\) | \(-1\) | \(3\) |
\(x\) | \(-1\) | \(3\) | \(5\) | \(1\) |
Vậy \(x\in\left\{1;-1;3;-3;5\right\}\)
Chúc bạn học tốt ~
tìm x: a)x^4-2x^3+5x^2-10x=0
b)(3x+5)^2=(2x-2)^2
. c)x^3–2x^2+x=0
. d)x^2(x-1)-4x^2+8x-4=0
\(a,x^4-2x^3+5x^2-10x=0\\ \Leftrightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Leftrightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x^2+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x\in\varnothing\left(x^2+5>0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(b,\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0\\ \Leftrightarrow\left(3x+5+2x-2\right)\left(3x+5-2x+2\right)=0\\ \Leftrightarrow\left(5x+3\right)\left(x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-7\end{matrix}\right.\)
\(c,x^3-2x^2+x=0\\ \Leftrightarrow x\left(x-1\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(d,x^2\left(x-1\right)-4x^2+8x-4=0\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a) \(x^4-2x^3+5x^2-10x=0\\ \Rightarrow\left(x^4-2x^3\right)+\left(5x^2-10x\right)=0\\ \Rightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Rightarrow\left(x^3+5x\right)\left(x-2\right)=0\\ \Rightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2+5=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{5}\\x=2\end{matrix}\right.\)
Vậy \(x=\left\{-\sqrt{5};0;\sqrt{5};2\right\}\)
b) \(\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Rightarrow\left[{}\begin{matrix}3x+5=2x-2\\3x+5=-2x+2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)
c) \(x^3-2x^2+x=0\\ \Rightarrow x\left(x^2-2x+1\right)=0\\ \Rightarrow x\left(x-1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
vậy ...
d) \(x^2\left(x-1\right)-4x^2+8x-4=0\\ x^2\left(x-1\right)-\left(4x^2-8x+4\right)=0\\ x^2\left(x-1\right)-\left(2x-2\right)^2=0\\ \Rightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\\ \Rightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Rightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a: Ta có: \(x^4-2x^3+5x^2-10x=0\)
\(\Leftrightarrow x\left(x^3-2x^2+5x-10\right)=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
b:Ta có: \(\left(3x+5\right)^2=\left(2x-2\right)^2\)
\(\Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0\)
\(\Leftrightarrow\left(3x+5-2x+2\right)\left(3x+5+2x-2\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(5x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)
tìm x biết
a/ x^3-x^2-x+1=0
b/(2x^3-3)^2-(4x^2-9)=0
c/x^4+2x^3-6x-9=0
d/2(x+5)-x^2-5x=0
1Rút gọn biểu thức a) (3x+1)^2+(3x-1)^2-2(3x+1)(3x-1) b) 8(3^2+1)(3^4+1)...(2^16+1) c ) (2^2+1)(2^4+1)...(2^32+1) 2 Tìm x biết a) x(2x-1)-2x+1=0 b) 3x(x-1)=x-1 c) 3(x+2)-x^2-2x=0 d) x^3+x=0 3 Phân tích thành nhân tử a) 4x^3-x b) 6x^2-12xy+6y^2-24z^2
Bài 2:
a: Ta có: \(x\left(2x-1\right)-2x+1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
Bài 1: (2đ). Thực hiện phép tính: a) 3x(x² + 2x - 1) b) (2x² +5x+2) : (x+2) 6 3 c) x² + 4x + 2x+8 Bài 2: (2đ). a) Tim x, biết: x(x – 2)+x−2 =0 a) x²-25-(x + 5) = 0 a) 2x²(3x² - 7x +2) b) (2x²-7x+3): (2x - 1) r 4-4x c) + x-2 x-2 x +1 -2x + c) 2x-2x² b) Tính giá trị của biểu thức: xẻ + 2x + l − y, tại x = 94,5 và y=4,5 b) Tính giá trị của biểu thức: (X + 1) − y”, tại x =94,5 và y=4,5 c) Tính giá trị biểu thức: Q = xẻ − 10x + 25 tại x = 1005 Bài 3: (2đ) Rút gọn phân thức a) A = x² +6x+9 b) 4x+10 2x²+5x B = c) C= x²-xy Sy²-5xy Bài 5: (2,5 đ) Cho AABC, đường trung tuyển AM. Gọi D là trung điểm của AB, E là điểm dối xứng với M qua D. a) Tử giác AEBM là hình gì? Vì sao? b) Biết AC = 12cm, tính độ dải đoạn MD?
Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
19 Tìm x, biết
a) (x+2)(x+3)-(x-2)(x+5)=0 ; b) (2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)
c) (8-4x)(x+2)+4(x-2)(x+1)=0 ; d) (2x-3)(8x+2)=(4x+1)(4x-1)-3
A. \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x^2+3x+2x+6\right)-\left(x^2+5x-2x-10\right)=0\)
\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)
\(\Leftrightarrow x^2+3x+2x-x^2-5x+2x=-6-10\)
\(\Leftrightarrow2x=-16\)
\(\Leftrightarrow x=-8\) .Vậy \(S=\left\{-8\right\}\)
B. \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x+5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x+5x-20\)
\(\Leftrightarrow2x^2-8x+3x+x^2-2x-5x-3x^2+12x-5x=12-10-20\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\) . Vậy \(S=\left\{\dfrac{18}{5}\right\}\)
C. \(\left(8-4x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow8x+16-4x^2-8x+4\left(x^2+x-2x-2\right)=0\)
\(\Leftrightarrow8x+16-4x^2-8x+4x^2+4x-8x-8=0\)
\(\Leftrightarrow8x-4x^2-8x+4x^2+4x-8x=-16+8\)
\(\Leftrightarrow-4x=-8\)
\(\Leftrightarrow x=2\) . Vậy \(S=\left\{2\right\}\)
D. \(\left(2x-3\right)\left(8x+2\right)=\left(4x+1\right)\left(4x-1\right)-3\)
\(\Leftrightarrow16x^2+4x-24x-6=16x^2+1^2-3\)
\(\Leftrightarrow16x^2+4x-24x-16x^2=6+1-3\)
\(\Leftrightarrow-20x=4\)
\(\Leftrightarrow x=-\dfrac{1}{5}\) . Vậy \(S=\left\{-\dfrac{1}{5}\right\}\)
a)(x+2)(x+3)-(x-2)(x+5)=0
\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)
<=>2x=-16
<=>x=-8
b)(2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)
\(\Leftrightarrow5x=22\Leftrightarrow x=\dfrac{22}{5}\)
c)(8-4x)(x+2)+4(x-2)(x+1)=0
\(\Leftrightarrow8x+16-4x^2-8x+4x^2+4x-8x-8=0\)
\(\Leftrightarrow-4x=-8\Leftrightarrow x=2\)
d)(2x-3)(8x+2)=(4x+1)(4x-1)-3
\(\Leftrightarrow16x^2+4x-24x-6=16x^2-4x+4x-1-3\)
\(\Leftrightarrow-20x=-2\Leftrightarrow x=\dfrac{-1}{10}\)
Tìm x,biết
a) ( x+2)×(x+3)-(x -2)×(x+5)=0
b) (2x+3)×(x-4)+(x-5)×(x-2)=(3x-5)×(x-4)
c) (8-5x)×(x+2)+4(x-2)×(x+1)+2(x-2)×(x+2)=0
d) (8x-3)×(3x+2)-(4x+7)×(x+4)=(2x+1)×(5x-1)-33
câu 1 Phân tích các đa thức sau thành nhân tử a, (1+2x).(1-2x) -x(x+2)(x-2) ; b, x^2+y^2-x^2y^2+xy-x-y; c, 2x^2(x+1)-x+1; câu 2 : Tìm x, biết : a, x^3-36x=0; b, ( 3x-1)^2=( x+3)^2; c, x^2(x-1)-4x^2+8x-4=0