1. (2x^2-3x+1)(2x^2+5x+1)-9x^2=0
d) (3x – 5)(7 – 5x) – (5x + 2)(2 – 3x) = 4 g) 3(2x - 1)(3x - 1) - (2x - 3)(9x - 1) =0 j) (2x – 1)(3x + 1) – (4 – 3x)(3 – 2x) = 3 k) (2x + 1)(x + 3) – (x – 5)(7 + 2x) = 8 m) 2(3x – 1)(2x + 5) – 6(2x – 1)(x + 2) = - 6
g: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
a) \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)
b) \(2x^3-5x^2+3x=0\)
c) \(9x^2-16-x\left(3x+4\right)=0\)
d) \(\dfrac{5x+4}{3}-1=\dfrac{3x-2}{4}\)
a) \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)\left(5x+8\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1-5x-8\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(-2x-7\right)=0\)
\(TH_1:3x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
\(TH_2:-2x-7=0\)
\(\Leftrightarrow x=-\dfrac{7}{2}\)
Vậy pt có tập nghiệm \(S=\left\{\dfrac{1}{3};-\dfrac{7}{2}\right\}\)
b) \(2x^3-5x^2+3x=0\)
\(\Leftrightarrow2x^3-2x^2-3x^2+3x=0\)
\(\Leftrightarrow2x^2\left(x-1\right)-3x\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)
\(TH_1:x=0\)
\(TH_2:x-1=0\)
\(\Leftrightarrow x=1\)
\(TH_3:2x-3=0\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy pt có tập nghiệm \(S=\left\{0;1;\dfrac{3}{2}\right\}\)
c) \(9x^2-16-x\left(3x+4\right)=0\)
\(\Leftrightarrow\left(9x^2-16\right)-x\left(3x+4\right)=0\)
\(\Leftrightarrow\left(3x-4\right)\left(3x+4\right)-x\left(3x+4\right)=0\)
\(\Leftrightarrow\left(3x+4\right)\left(2x-4\right)=0\)
\(TH_1:3x+4=0\)
\(\Leftrightarrow x=-\dfrac{4}{3}\)
\(TH_2:2x-4=0\)
\(\Leftrightarrow x=2\)
Vậy pt có tập nghiệm \(S=\left\{-\dfrac{4}{3};2\right\}\)
d) \(\dfrac{5x+4}{3}-1=\dfrac{3x-2}{4}\)
\(\Leftrightarrow\dfrac{20x+16}{12}-\dfrac{12}{12}=\dfrac{9x-6}{12}\)
\(\Rightarrow20x+16-12=9x-6\)
\(\Leftrightarrow20x-9x=-6-16+12\)
\(\Leftrightarrow11x=-10\)
\(\Leftrightarrow x=-\dfrac{10}{11}\)
Vậy pt có nghiệm duy nhất \(x=-\dfrac{10}{11}\)
a) Ta có: \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)=\left(3x-1\right)\left(5x+8\right)\)
\(\Leftrightarrow3x+1=5x+8\)
\(\Leftrightarrow3x-5x=8-1\)
\(\Leftrightarrow-2x=7\)
\(\Leftrightarrow x=\dfrac{-7}{2}\)
Vậy \(X=\dfrac{-7}{2}\)
b) Ta có: \(2x^3-5x^2+3x=0\)
\(\Leftrightarrow x\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x\left[\left(2x^2-2x\right)-\left(3x-3\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=0\) hoặc \(x=\dfrac{3}{2}\)
c) \(9x^2-16-x\left(3x+4\right)=0\)
\(\Leftrightarrow9x^2-16-3x^2-4x=0\)
\(\Leftrightarrow6x^2-4x-16=0\)
\(\Leftrightarrow2\left(3x^2-2x-8\right)=0\)
\(\Leftrightarrow3x^2-6x+4x-8=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-4}{3}\end{matrix}\right.\)
Vậy \(x=2\) hoặc \(x=\dfrac{-4}{3}\)
d) \(\dfrac{5x+4}{3}-1=\dfrac{3x-2}{4}\)
\(\Leftrightarrow\dfrac{20x+16}{12}-\dfrac{12}{12}=\dfrac{9x-6}{12}\)
\(\Leftrightarrow20x+16-12=9x-6\)
\(\Leftrightarrow20x+16-12-9x+6=0\)
\(\Leftrightarrow11x+10=0\)
\(\Leftrightarrow x=\dfrac{-10}{11}\)
Vậy \(x=\dfrac{-10}{11}\)
a) Ta có: \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)\left(5x+8\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1-5x-8\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(-2x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\-2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\-2x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{1}{3};-\dfrac{7}{2}\right\}\)
a) -3x(2x-5)-2x(2-3x)=7
b)(9x-12x+4)(2-5x)=0
c)(4-3x)=(5+2x)
d)(2x-1)-3(2x-1)=0
a) -3x(2x-5)-2x(2-3x)=7
=> -6x2 + 15 - 4x + 6x2 = 7
=> -6x2 + 6x2 + 15 -4x =7
=> 15 - 4x =7
=> 4x = 15-7 =8
=> x= 8:4 = 2
b) \(\left(9x-12x+4\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}9x-12x+4=0\\2-5x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(9-12\right)=-4\\5x=2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}-3x=-4\\x=\frac{2}{5}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{2}{5}\end{cases}}\)
Vay...
c) (4-3x) = (5+2x)
=> 4-3x=5+2x
=> -3x - 2x = 5-4
=> x(-3-2) = 1
=> -5x = 1
=> x= \(\frac{-1}{5}\)
d) (2x-1)-3(2x-1)=0
=> 2x-1 - 6x + 3 =0
=> 2x - 6x = 1 -3
=> x(2-6)=-2
=> -4x= -2
=> x = \(\frac{1}{2}\)
d)(2x-1)-3(2x-1)
=>1(2x-1)-3(2x-1)=0
=>(1-3).(2x-1)=0
=>-2(2x-1)=0
=>2x-1=0
=>2x=-1
=>x=-0,5
vay x =-0,5
giải các phương trình sau
a.3(x-1)=5x+8
b.9x^2-1=(3x+1)(4x+1)
c.(2x+1)^2=(x-1)^2
d.2x^3+3x^3-5x=0
e.x^2+2x-15=0
a) \(3\left(x-1\right)=5x+8\)
\(\Leftrightarrow\)\(3x-3=5x+8\)
\(\Leftrightarrow\)\(2x=-11\)
\(\Leftrightarrow\)\(x=-5,5\)
Vậy...
b) \(9x^2-1=\left(3x+1\right)\left(4x+1\right)\)
\(\Leftrightarrow\)\(\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\)\(\left(3x+1\right)\left(3x-1-4x-1\right)=0\)
\(\Leftrightarrow\)\(\left(3x+1\right)\left(-x-2\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+1=0\\-x-2=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{1}{3}\\x=-2\end{cases}}\)
Vậy..
c) \(\left(2x+1\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow\)\(\left(2x+1\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\)\(\left(2x+1-x+1\right)\left(2x+1+x-1\right)=0\)
\(\Leftrightarrow\)\(3x\left(x+2\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x+2=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
Vậy...
d) \(2x^3+3x^3-5x=0\)
\(\Leftrightarrow\)\(5x^3-5x=0\)
\(\Leftrightarrow\)\(5x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\)\(x=0\)hoặc \(x-1=0\)hoặc \(x+1=0\)
\(\Leftrightarrow\)\(x=0\) hoặc \(x=1\) hoặc \(x=-1\)
Vậy...
p/s: chỗ "hoặc" bn đưa về kí hiệu "[" cho mk nhé
e) \(x^2+2x-15=0\)
\(\Leftrightarrow\)\(\left(x-3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-3=0\\x+5=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)
Vậy...
a,\(\Leftrightarrow3x-3=5x+8\)
\(\Leftrightarrow2x=-11\)
\(\Leftrightarrow x=-\frac{11}{2}\)
b,\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)=\left(3x+1\right)\left(4x+1\right)\)=0
\(\Leftrightarrow\left(3x+1\right)\left(3x+1-4x-1\right)\)=0
\(\Leftrightarrow\left(3x+1\right)\left(-x\right)\)=0
\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{3}\\x=0\end{cases}}\)
c\(\Leftrightarrow4x^2+4x+1=x^2-2x+1\)
\(\Leftrightarrow3x^2+6x=0\)
\(\Leftrightarrow3x\left(x+2\right)\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
d,có lẽ bạn viết sai đề phải ko
2x3+3x2-5x=0
\(\Leftrightarrow x\left(2x^2+3x-5\right)=0\)
\(\Leftrightarrow x\left(2x^2-2x+5x-5\right)\Leftrightarrow x\left(x-1\right)\left(2x+5\right)\)
\(\orbr{\begin{cases}x=0\\x-1=0\end{cases}}va.2x+5=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{2}\\x=0.và.x=1\end{cases}}\)
e,\(x^2+2x-15=0\)
\(\Leftrightarrow x^2-3x+5x-15\)=0
\(\Leftrightarrow\left(x-3\right)\left(x+5\right)\)=0
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)
Tìm x, biết:
a)2x*(6x-5)-4x*(3x+7)=7
b)-5x(2x+1)+3x(3x+2)+x(x-1/2)=0
c)3x*(6x-5)-2x(9x+7)=15
d)1/2x*(2x+4)-(x+3)=5
e)(-3x+2)*5x-5x*(2x+1)-5x=4
Mấy bạn giúp mk ik mk đang cần gấp!
a: \(\Leftrightarrow12x^2-10x-12x^2-28x=7\)
=>-38x=7
hay x=-7/38
b: \(\Leftrightarrow-10x^2-5x+9x^2+6x+x^2-\dfrac{1}{2}x=0\)
=>1/2x=0
hay x=0
c: \(\Leftrightarrow18x^2-15x-18x^2-14x=15\)
=>-29x=15
hay x=-15/29
d: \(\Leftrightarrow x^2+2x-x-3=5\)
\(\Leftrightarrow x^2+x-8=0\)
\(\text{Δ}=1^2-4\cdot1\cdot\left(-8\right)=33>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{33}}{2}\\x_2=\dfrac{-1+\sqrt{33}}{2}\end{matrix}\right.\)
e: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)
\(\Leftrightarrow-25x^2=4\)
\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)
1) (4-3x) (10x-5)=0
2) (7-2x) (4+8x) = 0
3) (9-7x) (11-3x) = 0
4) (7-14x) (x-2) = 0
5) (2x+1) (x-3) = 0
6) (8-3x) (-3x+5) = 0
7) (16-8x) (2-6x) = 0
8) (x+4) (6x-12) = 0
9) (11-33x) (x+11) = 0
10) (x-1/4) (x+5/6) = 0
11) (7/8-2x) (3x+1/3) = 0
12) 3x - 2x^2 = 0
13) 5x + 10x^2 = 0
14) 4x + 3x^2 = 0
15) -8x^2 + x =0
16) 10x^2 - 15x = 0
17) x^2 -4 =0
18) 9 - x^2 = 0
19) x^2 -1 = 0
20) (x-3) (2x-1) = (2x-1) ( 2x+3)
21) (5+4x) (-x+2) = (5+4x) (7+5x)
22) (4+x) (x-5) = (3x-8) (x-5) = 0
23) (3x-8) (7-21x) - (9+2x) (7-21x)
24) (10+ 7x) (x+1) = (9x-2)(x-1)
25) (9x-4) (x-1/2) - (x-1/2) (6+x) = 0
26) 9x^2 - 1 = (3x-1) (x+4)
27) (x+7) (3x+1) = 49-x^2
28) (2x+1)^2 = (x-1)^2
29)x^3- 5x^2+6x = 0
30) 3x^2 + 5x + 2 = 0
Giảii giúpp mìnhh đyy mọii ngườii .
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha
Tìm x:
a)9x^2-30x+25=0
b)25x^2-5x+1/4=0
c)9x^2-25=0
d)(2x-1)^2-(3x+2)^2=0
a: \(9x^2-30x+25=0\)
\(\Leftrightarrow3x-5=0\)
hay \(x=\dfrac{5}{3}\)
c: \(9x^2-25=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
a) \(9x^2-30x+25=0\Rightarrow\left(3x-5\right)^2=0\Rightarrow x=\dfrac{5}{3}\)
b) \(25x^2-5x+\dfrac{1}{4}=0\Rightarrow\left(10x-1\right)^2=0\Rightarrow x=\dfrac{1}{10}\)
c) \(9x^2-25=0\Rightarrow\left(3x-5\right)\left(3x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
d) \(\left(2x-1\right)^2-\left(3x+2\right)^2=0\)
\(\Rightarrow\left(2x-1+3x+2\right)\left(2x-1-3x-2\right)=0\)
\(\Rightarrow-\left(5x+1\right)\left(5x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
tim x biết
3x+4=0
2x*(x-1)-(1+2x)=-34
X^2+9x-10=0
(7x-1)*(2+5x)=0
\(3x+4=0\Leftrightarrow x=-\dfrac{4}{3}\\ 2x\left(x-1\right)-\left(1+2x\right)=-34\\ \Leftrightarrow2x^2-2x-1-2x=-34\\ \Leftrightarrow2x^2-4x+33=0\\ \Leftrightarrow2\left(x^2-2x+1\right)+30=0\\ \Leftrightarrow2\left(x-1\right)^2+30=0\\ \Leftrightarrow x\in\varnothing\left[2\left(x-1\right)^2+30\ge30>0\right]\\ x^2+9x-10=0\\ \Leftrightarrow x^2-x+10x-10=0\\ \Leftrightarrow\left(x-1\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-10\end{matrix}\right.\\ \left(7x-1\right)\left(2+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}7x-1=0\\2+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
Bài:Chia 1 biến đã sắp xếp 1)(2x^3+11x^2+18x-3):(2x+3) 2)(2x^3+11x^2+18x-3):(3x+3) 3)(2x^3+9x^2+5x+41):(2x^2-x+9) 4)(13x+41x^2+35x^3-14):(5x-2) 5)(5x^2-3x^3+15-9x):(5-3x) 6)(-4x^2+x^3-20+5x):(x-4)
1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)
\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)
\(=x^2+4x+3-\dfrac{12}{2x+3}\)