a.(x+5)(2x-1)=(2x-3)(x+1)
b.(x+1)(x+9)=(x+3)(x+5)
c.(3x+5)(2x+1)=(6x-2)(x-3)
d.(x-2)(3x+5)=(2x-4)(x+1)
đ.9x2 -1=(3x+1)(2x-3)
e.(2x+5)(x-4)=(x-5)(4-x)
a,(3x - 1)(x + 3) = (2 - x)(5 - 3x)
b,(x + 5)(2x - 1) = (2x - 3)(x + 1)
c,(x + 1)(x + 9) = (x + 3)(x + 5)
d,(3x + 5)(2x + 1) = (6x - 2)(x - 3)
e,(x + 2)2 + 2(x - 4) = (x - 4)(x - 2)
f,(x + 1)(2x - 3)-(3x - 2) = 2(x - 1)2
a) \(\left(3x-1\right)\left(x+3\right)=\left(2-x\right)\left(5-3x\right)\)
\(\Leftrightarrow3x^2+8x-3=3x^2-11x+10\)
\(\Leftrightarrow19x-13=0\)
\(\Leftrightarrow x=\frac{13}{19}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{13}{19}\right\}\)
b) \(\left(x+5\right)\left(2x-1\right)=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow2x^2+9x-5=2x^2-x-3\)
\(\Leftrightarrow10x-2=0\)
\(\Leftrightarrow x=\frac{1}{5}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{1}{5}\right\}\)
c) \(\left(x+1\right)\left(x+9\right)=\left(x+3\right)\left(x+5\right)\)
\(\Leftrightarrow x^2+10x+9=x^2+8x+15\)
\(\Leftrightarrow2x-6=0\)
\(\Leftrightarrow x=3\)
Vậy tập nghiệm của phương trình là \(S=\left\{3\right\}\)
d) \(\left(3x+5\right)\left(2x+1\right)=\left(6x-2\right)\left(x-3\right)\)
\(\Leftrightarrow6x^2+13x+5=6x^2-20x+6\)
\(\Leftrightarrow33x-1=0\)
\(\Leftrightarrow x=\frac{1}{33}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{1}{33}\right\}\)
e) \(\left(x+2\right)^2+2\left(x-4\right)=\left(x-4\right)\left(x-2\right)\)
\(\Leftrightarrow x^2+4x+4+2x-8=x^2-6x+8\)
\(\Leftrightarrow6x-4=-6x+8\)
\(\Leftrightarrow12x-12=0\)
\(\Leftrightarrow x=1\)
Vậy tập nghiệm của phương trình là \(S=\left\{1\right\}\)
f) \(\left(x+1\right)\left(2x-3\right)-\left(3x-2\right)=2\left(x-1\right)^2\)
\(\Leftrightarrow2x^2-x-3-3x+2=2\left(x^2-2x+1\right)\)
\(\Leftrightarrow2x^2-4x-1=2x^2-4x+2\)
\(\Leftrightarrow-1=2\)(ktm)
Vậy tập nghiệm của phương trình là \(S=\varnothing\)
Giải :
a) \(\left(3x-1\right)\left(x+3\right)=\left(2-x\right)\left(5-3x\right)\)
\(\leftrightarrow3x^2+8x-3-10+11x-3x^2=0\)
\(\leftrightarrow19x-13=0\)
\(\leftrightarrow x=\frac{13}{19}\)
Vậy phương trình có nghiệm là \(x=\frac{13}{19}\)
b) \(\left(x+5\right)\left(2x-1\right)=\left(2x-3\right)\left(x+1\right)\)
\(\leftrightarrow2x^2+9x-5-2x^2+x+3=0\)
\(\leftrightarrow10x-2=0\)
\(\leftrightarrow x=\frac{1}{5}\)
Vậy phương trình có nghiệm là \(x=\frac{1}{5}\)
c) \(\left(x+1\right)\left(x+9\right)=\left(x+3\right)\left(x+5\right)\)
\(\leftrightarrow x^2+10x+9-x^2-8x-15=0\)
\(\leftrightarrow2x-6=0\)
\(\leftrightarrow x=3\)
Vậy phương trình có nghiệm là \(x=3\)
d) \(\left(3x+5\right)\left(2x+1\right)=\left(6x-2\right)\left(x-3\right)\)
\(\leftrightarrow6x^2+13x+5-6x^2+20x-6=0\)
\(\leftrightarrow33x-1=0\)
\(\leftrightarrow x=\frac{1}{33}\)
Vậy phương trình có nghiệm là \(x=\frac{1}{33}\)
e) \(\left(x+2\right)^2+2\left(x-4\right)=\left(x-4\right)\left(x-2\right)\)
\(\leftrightarrow x^2+4x+4+2x-8-x^2+6x-8=0\)
\(\leftrightarrow12x-12=0\)
\(\leftrightarrow x=1\)
Vậy phương trình có nghiệm là \(x=1\)
f) \(\left(x+1\right)\left(2x-3\right)-\left(3x-2\right)=2\left(x-1\right)^2\)
\(\leftrightarrow2x^2-x-3-3x+2-2x^2+4x-2=0\)
\(\leftrightarrow-3=0\left(VL\right)\)
Vậy phương trình này vô nghiệm
Nhớ k mik nhe , mik camon cậu
Bài 4: Tìm x, biết:
a) 3(2x – 3) + 2(2 – x) = –3 ; b) x(5 – 2x) + 2x(x – 1) = 13 ;
c) 5x(x – 1) – (x + 2)(5x – 7) = 6 ; d) 3x(2x + 3) – (2x + 5)(3x – 2) = 8 ;
e) 2(5x – 8) – 3(4x – 5) = 4(3x – 4) + 11; f) 2x(6x – 2x 2 ) + 3x 2 (x – 4) = 8.
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy: \(x=\dfrac{1}{2}\)
===========
b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
\(\Leftrightarrow x=\dfrac{13}{3}\)
Vậy: \(x=\dfrac{13}{3}\)
==========
c/ \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)
\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)
\(\Leftrightarrow-2x=-2\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
\(\Leftrightarrow x=\dfrac{2}{7}\)
Vậy: \(x=\dfrac{2}{7}\)
==========
f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow-x^3=8\)
\(\Leftrightarrow x=-2\)
Vậy: \(x=-2\)
Tìm x
a.(x+2).(x+3)-(x-2).(x+5) = 0
b. (2x+3).(x-4)+(x-5)(x+2) = (3x-5)(x-4)
c. (3x+2)(2x+9)-(x+2)(6x+1) = x+1-(x-6)
d. 3( 2x-1).(3x-1)-(2x-3).(9x-1)=0
(x+2)(x+3)-(x-2)(x+5)=0
=> x2+5x+6-x2-3x+10=0
=>2x+16=0
=>2x=-16
=>x=-8
Bài 1: giải các phương trình sau
a,(3x-1)(x+3)=(2-x)(5-3x) d,(3x+5)(2x+1)=(6x-2)(x-3)
b,(x+5)(2x-1)=(2x-3)(x+1) e,(x+2)2+2(x-4)=(x-4)(x-2)
c,(x+1)(x+9)=(x+3)(x+5) f,(x+1)(2x-3)-3(x-2)=2(x-1)2
\(c.\left(x+1\right)\left(x+9\right)=\left(x+3\right)\left(x+5\right)\\\Leftrightarrow x^2+9x+x+9=x^2+5x+3x+15\\\Leftrightarrow x^2-x^2+9x+x-5x-3x=-9+15\\\Leftrightarrow 2x=6\\\Leftrightarrow x=3\)
Vậy nghiệm của phương trình trên là \(3\)
\(f.\left(x+1\right)\left(2x-3\right)-3\left(x-2\right)=2\left(x-1\right)^2\\\Leftrightarrow 2x^2-3x+2x-2-3x+6=2x^2-4x+2\\\Leftrightarrow 2x^2-2x^2-3x+2x-3x+4x=2-6+1\\ \Leftrightarrow0x=-3\)
\(\Rightarrow\) Vô nghiệm
\(a.\left(3x-1\right)\left(x+3\right)=\left(2-x\right)\left(5-3x\right)\\\Leftrightarrow 3x^2+9x-x-3=10-6x-5x+3x^2\\\Leftrightarrow 3x^2-3x^2+9x-x+6x+5x=3+10\\ \Leftrightarrow19x=13\\\Leftrightarrow x=\frac{13}{19}\)
Vậy nghiệm của phương trình trên là \(\frac{13}{19}\)
Tìm x biết :
a, 4.(18 - 5x) - 12.(3x - 7) = 15.(2x - 16) - 6(x + 14)
b, 5.(3x + 5) - 4.(2x - 3) = 5x + 3.(2x + 12) + 1
c, 2.(5x - 8) - 3.(4x - 5) = 4.(3x - 4) + 11
d, (3x + 2)(2x + 9) - (x + 2)(6x + 1) = (x + 1) - (x - 6)
e, (8x - 3)(3x + 2) - (4x + 7)(x + 4)= (2x + 1)(5x - 1) - 33
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
b, 5(3x + 5) - 4(2x - 3) = 5x + 3(2x + 12) + 1
=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1
=> (15x - 8x) + (25 + 12) = 11x + 37
=> 7x + 37 = 11x + 37
=> 11x - 7x = 0
=> x = 0
Bài 1 : Tìm x, biết
a) 3x(x2 - 2) + x2(1-2x) - x2 = -12
b) 3x(2x +3) - (2x +5)(3x -2) = 8
c)3(3x -1)(2x+5) - 6(2x -1)(x+2) = -6
d) 3(2x -1)(3x - 1)- (2x -3)(9x -1) - 3 = -3
e) 4x(x-1)- 3(x2 -5)- x2 = (x-3)(x+4)
f) (3x-1)(2x+7)-(x+1)(6x-5)= (x+2)-(x-5)
tck đầu tiên chọn câu trả lời của mình đi
a. (2x + 5)(2x - 5)
b. (x2 + 3)(3 - x2 )
c. 3x(x -1)2 - 2x(x + 3)(x - 3) + 4x(x - 4)
d. 3x(4x - 3) - (2x -1)(6x + 5)
e. 4x(3x2 - x) - (2x + 3)(6x2 - 3x + 1)
f. (x - 2)(1x + 2)(x + 4)
\(\left(x^2+3\right)\left(3-x^2\right)\)
\(\left(x^2+3\right)\left(-x^2+3\right)\)
\(\left(-x^2+3\right).x^2+3\left(-x^2+3\right)\)
\(-x^2.x^2+3x^2+3\left(-x^2+3\right)\)
\(-x^2.x^2+3x^2-3x^2+9\)
\(-x^2.x^2+9\)
\(\left(2x+5\right)\left(2x-5\right)\)
\(2x\left(2x-5\right)+5\left(2x-5\right)\)
\(4x^2-10x+5\left(2x-5\right)\)
\(4x^2-10x+10x-25\)
\(4x^2-25\)
(x+2)(x+3)-(x-2)(x+5)=0
b)(3x-1)(2x+7)-(x+1(6x-5)
c)(10x+9x)x-(5x-1)(2x+30=8
d)x(2x-1)(x+5)-(2x2+1)(x+4,5)=3,5
e)(2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)
làm phép chia :
a) (x^4 -2x^3 + 2x -1) : (x^2 - 1)
b) (x^3 -8) : (x^2 + 2x +4)
c) (x^6 - 2x^5 + 2x^4 + 6x^3 - 4x^2)n: 6x^2
d) (-2x^5 + 3x^2 - 4x^3) :2x^2
e) (15x^3 - 10x^2 + x - 2) : (x - 2)
f) (2x^4 - 3x^3 - 3x^2 + 6x - 2) : (x^2 - 2)
b: =x-2
d: \(=-x^3+\dfrac{3}{2}-2x\)