Tìm x biết
a) 3(1-4x).(x-1)+4(3x-2).(x+3)=-27
b) (x+3).(x2-3+9)-.(x-1).(x+1)=27
c) 2x2+3(x-1).(x+1)=5x.(x+1)
d) 8(x-1).(x+5)-(x+2).(x+5)=3(x-1).(x+2)
Bài 1:Thực hiện phép tính
a,(5-2x)(x+3)-4x(x+2) b,(3x+1)(x-3)-4(x+2)(x-2)
c,3(x-4)(x+3)+(x-5)(x+3) d,2x(x-4)+(3x-1)(2x-5)
Bài 2:Tìm x biết
a,5x(x+3)-(5x+2)(x+3)=7
b,(3x-1)(3x+2)-9(x+2)(x-2)=10
c,(x+1)(2x-5)+2(3-x)(x+2)=7
d,(1-3x)(x+2)+3x(x-5)=8
Bài 5: Tìm nghiệm của các đa thức sau: Dạng 1: a) 4x + 9 b) -5x + 6 c) 7 – 2x d) 2x + 5 Dạng 2: a) ( x+ 5 ) ( x – 3) b) ( 2x – 6) ( x – 3) c) ( x – 2) ( 4x + 10 ) Dạng 3: a) x2 -2x b) x2 – 3x c) 3x2 – 4x d) ( 2x- 1)2 Dạng 4: a) x2 – 1 b) x2 – 9 c)– x 2 + 25 d) x2 - 2 e) 4x2 + 5 f) –x 2 – 16 g) - 4x4 – 25 Dạng 5: a) 2x2 – 5x + 3 b) 4x2 + 6x – 1 c) 2x2 + x – 1 d) 3x2 + 2x – 1
tìm x biết
a,5x(x-4)3(x+2)(x-4)=2x(x+1)
b,4x(x+2)-x(8x-5)=10
c,(x+3)(2x-5)=2x(x+4)
d,(3x-2)(x+5)-3x(x+4)=5
e,x(x-3)+2x(x+1)=3(x mũ2-4)
b: =>4x^2+8x-8x^2+5x-10=0
=>-4x^2+13x-10=0
=>x=2 hoặc x=5/4
c: =>2x^2-5x+6x-15=2x^2+8x
=>x-15=8x
=>-7x=15
=>x=-15/7
d: =>3x^2+15x-2x-10-3x^2-12x=5
=>x-10=5
=>x=15
e: =>x^2-3x+2x^2+2x=3x^2-12
=>-x=-12
=>x=12
a. (x – 1)(5x + 3) = (3x – 8)(x – 1)
b. 3x(25x + 15) – 35(5x + 3) = 0
c. (2 – 3x)(x + 11) = (3x – 2)(2 – 5x)
d. (2x2 + 1)(4x – 3) = (2x2 + 1)(x – 12)
e. (2x – 1)2 + (2 – x)(2x – 1) = 0
f. (x + 2)(3 – 4x) = x2 + 4x + 4
\(a,\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
\(\left(x-1\right)\left(5x+3-3x+8\right)=0\)
\(\left(x-1\right)\left(2x+11\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\2x=-11\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-\frac{11}{2}\end{cases}}}\)
\(b,3x\left(25x+15\right)-35\left(5x+3\right)=0\)
\(15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\left(5x+3\right).5\left(3x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x+3=0\\5\left(3x-7\right)=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-3\\3x-7=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\x=\frac{7}{3}\end{cases}}}\)
\(c,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)
\(\left(3x-2\right)\left(2-5x\right)+\left(3x-2\right)\left(x+11\right)=0\)
\(\left(3x-2\right)\left(2-5x+x+11\right)=0\)
\(\left(3x-2\right)\left(13-4x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-2=0\\13-4x=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=2\\4x=13\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{13}{4}\end{cases}}}\)
còn đâu tự lm lười :_#
tìm x , biết
a. 4x(x-5)-(x-1)(4x-3)=5
b. (3x-4)(x-2) = 3x(x-9)-3
c.2(x+3)-x2 -3x=0
d. 8x3-50x=0
e. (4x-30)2-3x(3-4x)
\(a,\Rightarrow4x^2-20x-4x^2+3x+4x-3=5\\ \Rightarrow-13x=8\Rightarrow x=-\dfrac{8}{13}\\ b,\Rightarrow3x^2-10x+8-3x^2+27x=-3\\ \Rightarrow17x=-11\Rightarrow x=-\dfrac{11}{17}\\ c,\Rightarrow\left(x+3\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ d,\Rightarrow2x\left(4x^2-25\right)=0\\ \Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\\ e,Sửa:\left(4x-3\right)^2-3x\left(3-4x\right)=0\\ \Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\\ \Rightarrow\left(4x-3\right)\left(7x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
a.
4x(x-5) - (x-1)(4x-3)-5=0
4x^2-20x-4x^2+3x+4x+3=0
(4x^2-4x^2)+(-20x+3x+4x)+3=0
13x+3 = 0
13x=-3
x=-3/13
b,
(3x-4)(x-2)-3x(x-9)+3=0
3x^2-6x-4x+8 - 3x^2+27x+3=0
(3x^2-3x^2)+(-6x-4x+27x)+(8+3)=0
17x+11=0
17x=-11
x=-11/17
c, 2(x+3)-x^2-3x=0
2(x+3) - x(x+3)=0
(x+3)(2-x)=0
TH1: x+3 = 0; x=-3
TH2: 2-x=0;x=2
1) √(2x-1) <= 8-2x
2) √[(x+1)(4-x)] > x-2
3) √(x-2x^2+1) > 1-x
4) √(x+5) - √(x+4) > √(x+3)
5) √(5x-1) - √(x-1) > √(2x-4)
6) √(x+3) >= √(2x-8) + √(7-x)
7) √(x+2) - √(3-x) < √(5-2x)
8) √(x+1) > 3 - √(x+4)
9) √(5x-1) - √(4x-1)<= 3√x
10) { {√[2(x^2-16)]} / √(x-3) }+ √(x-3) > (7-x) / √(x-3)
Giúp mình 10 câu này với ạaa
Bạn nên viết đề bằng công thức toán và ghi đầy đủ yêu cầu đề để mọi người hiểu đề của bạn hơn nhé.
Chủ đề 1: Thực hiện phép tính
1) (2x+3).(2x-3)-4x.(x+5)
2) 6/x2 - 9 + 5/x-3 + 1/x+3
3)5x.(x-3)+(x-2)2
4) 4x/x+2 - 3x/x-2 + 12x/ x2 - 4
5) x(x+2) - ( x-3)(x+3)
6) 1/3x-2 + -4/3+2 + 6-3x/9x2 - 4
7)2x.(3x-1)+(x+2)2
8) 6/x+3 - 6/x-3 + 9x+9/x2 - 9
9) (2x - 5)2 - x(4x-13)
10) x-1/x + 4/x+8 + 8/x2 + 8x
11) (2x+1)2 + (x-5)(x+5)-x(5x+7)
12) 6/x2-9 + 5/x-3 + 1/x+3
13) 6x(5x-2)+(2x+3)2
14) x/x-2 + -2/x-3 + x(1-x)/x2-9
15) (x-2)2-x(x+5)
16) 2/x+3 + 3/x-3 + -6/x2-9
17) 3x(x-3) + (3x-1)2
\(\left(2x+3\right)\left(2x-3\right)-4x\left(x+5\right)=4x^2-9-4x^2-20x=-20x-9\)
\(5x\left(x-3\right)+\left(x-2\right)^2=5x^2-15x+x^2-4x+4=6x^2-19x+4\)
\(x\left(x+2\right)-\left(x-3\right)\left(x+3\right)=x^2+2x-\left(x^2-9\right)=x^2+2x-x^2+9=2x+9\)
Bài 1: Thu gọn các biểu thức sau
a)(2x2 + 5x - 2)(2x2 - 4x +3)
b)(2x -3)(3x - 2) - 3x(2x - 5)
c)(x -1)(x2 + x + 1) - (x + 1)(x2 - x +1)
d)(x2 + x - 1)(x2 - x + 1)
e)(2 + 3y)2 - (2x -3y)2 -12xy
d)(x2 - 4x)(5 + 2x - x2)
cảm ơn!giúp mình với chiều nay ktra 15ph T_T
Bài 1 Rút gọn biểu thức
a, [(3x - 2)(x + 1) - (2x + 5)(x2 - 1)] : (x + 1)
b, (2x + 1)2 - 2(2x + 1)(3 - x) + (3 - x)2
c, (x - 1)2 - (x + 1) (x2 - x + 1) - (3x + 1)(1 - 3x)
d, (x2 + 1)(x - 3) - (x - 3)(x2 + 3x + 9)
e, (3x +2)2 + (3x - 2)2 - 2(3x + 2)(3x - 2) + x
Bài 2 Phân tích các đa thức sau thành nhân tử
1, 3(x + 4) - x2 - 4x
2, x2 - xy + x - y
3, 4x2 -25 + (2x + 7)(5 - 2x)
4, x2 + 4x - y2 + 4
5, x3 - x2 - x + 1
6, x3 + x2y - 4x - 4y
7, x3 - 3x2 + 1 - 3x
8, 2x2 + 3x - 5
9, x2 - 7xy + 10y2
10, x3 - 2x2 + x - xy2
1) 5(x-3) (x-7)-(5x+1) (x-2)= -8
2) x(x+1) (x+2)-(x+4) (3x-5)= 84-5x
3) (9x2-5) (x+3)-3x2(3x+9)=(x-5) (x+4)-x(x-11)
4) (x2-4x+16) (x+4)-x(x+1) (x+2)+3x2=0
5) (8x+2) (1-3x)+(6x-1) (4x-10)=-50
6) (x2+2x+4) (2-x)+x(x-3) (x+4)-x2+24=0
7) (\(\dfrac{x}{2}\)+3) (5-6x)+(12x-2) (\(\dfrac{x}{4}\)+3)=0
1) Ta có: \(5\left(x-3\right)\left(x-7\right)-\left(5x+1\right)\left(x-2\right)=-8\)
\(\Leftrightarrow5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=-8\)
\(\Leftrightarrow5x^2-50x+105-5x^2+9x+2+8=0\)
\(\Leftrightarrow-41x=-115\)
hay \(x=\dfrac{115}{41}\)
2) Ta có: \(x\left(x+1\right)\left(x+2\right)-\left(x+4\right)\left(3x-5\right)=84-5x\)
\(\Leftrightarrow x\left(x^2+3x+2\right)-\left(3x^2+7x-20\right)=84-5x\)
\(\Leftrightarrow x^3+3x^2+2x-3x^2-7x+20-84+5x=0\)
\(\Leftrightarrow x^3=64\)
hay x=4
3) Ta có: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)
\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)
\(\Leftrightarrow-5x-15=10x-20\)
\(\Leftrightarrow-5x-10x=-20+15\)
\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)