1) 5.(x-2)+27=4x-8
2) 3.(x+1)+2.(x+2)=-13
3)(2x+4).(5-x)=0
4)x2 -3x=0
5) x2 -3x+2=0
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: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)
\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)
\(\Leftrightarrow-5x-15=10x-20\)
\(\Leftrightarrow-5x-10x=-20+15\)
\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)
Bài 1: Giải các phương trình dưới đây
1) x2 - 9 = (x - 3)(5x +2)
2) x3 - 1 = (x - 1)(x2 - 2x +16)
3) 4x2 (x - 1) - x + 1 = 0
4) x3 + 4x2 - 9x - 36 = 0
5) (3x + 5)2 = (x - 1)2
6) 9 (2x + 1)2 = 4 (x - 5)2
7) x2 + 2x = 15
8) x4 + 5x3 + 4x2 = 0
9) (x2 - 4) - (x - 2)(3 - 2x) = 0
10) (3x + 2)(x2 - 1) = (9x2 - 4) (x + 1)
11) (3x - 1)(x2 + 2) = (3x - 1)(7x - 10)
12) (2x2 + 1) (4x - 3) = (x - 12)(2x2 + 1)
1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)
hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)
2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)
hay \(x\in\left\{1;5\right\}\)
3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)
\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)
\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)
hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)
1.
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)
\(\Leftrightarrow x+3=5x-2\)
\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)
2.
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)
\(\Leftrightarrow x^2+x+1=x^2-2x+16\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
3.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
7.
\(\Leftrightarrow x^2+2x-15=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
8.\(\Leftrightarrow x^4+x^3+4x^3+4x^2=0\)
\(\Leftrightarrow x^3\left(x+1\right)+4x^2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+4x^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0;x=-4\end{matrix}\right.\)
9.\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(3-2x\right)\)
\(\Leftrightarrow x+2=3-2x\)
\(\Leftrightarrow3x=1\Leftrightarrow x=\dfrac{1}{3}\)
1) (3x - 2)(4x + 5) = 0
2) (4x + 2)(x2 + 3) = 0
3) (2x + 7)(x - 3)(5x - 1) = 0
4) x2 - 3x = 0
5) x2 - x = 0
1
(3x-2)(4x+5)=0
⇔ 3x-2=0 -> x= 2/3
⇔ 4x-5=0 x= 5/4
Vậy tập nghiệm S = { 2/3; 5/4}
2, (4x+2)(\(X^2\)+3)=0
⇔ 4x+2=0 -> x= -1/2
\(x^2\)+3=0 -> x= \(\sqrt{3}\); -\(\sqrt{3}\)
Vaayj tập nghiệm S= { -1/2; \(\sqrt{3}\);-\(\sqrt{3}\)}
3)
(2x+7)(x-3)(5x-1)=0
⇔ 2x+7=0 -> x= -7/2
x-3 =0 -> x = 3
5x-1 =0 -> x= 1/5
Vậy tập nghiệm S={ -7/2; 3; 1/5}
1) x2 - x - (3x - 3) = 02) x(x - 6) - 7x + 42 = 0
3) x3 - 3x2 + 3x - 1 = 0
4) (2x - 5)2 - (x + 2) 2 = 0
5) x(2x - 9) = 3x(x - 5)
1) x^2-x-(3x-3)=0
⇔ X^2-x-3x+3=0
⇔ x^2-4x+3 =0
⇔x^2-3x-x+3 =0
⇔ x(x-3)-(x-3) =0
⇔(x-1)(x-3) =0
⇔ x-1=0 -> x=1
x-3=0 -> x=3
Vậy tập nghiệm S={ 1;3}
1) (x2-4x+16) (x+4)-x(x+1) (x+2)+3x2=0
2) (8x+2) (1-3x)+(6x-1) (4x-10)=-50
3) (x2+2x+4) (2-x)+x(x-3) (x+4)-x2+24=0
4) (\(\dfrac{x}{2}\)x2+3) (5-6x)+(12x-2) (\(\dfrac{x}{4}\)x4+3)=0
1)(x2-4x+16)(x+4)-x(x+1)(x+2)+3x2=0
\(\Rightarrow\)(x3+64)-x(x2+2x+x+2)+3x2=0
\(\Rightarrow\)x3+64-x3-2x2-x2-2x+3x2=0
\(\Rightarrow\)-2x+64=0
\(\Rightarrow\)-2x=-64
\(\Rightarrow\)x=\(\dfrac{-64}{-2}\)
\(\Rightarrow x=32\)
2)(8x+2)(1-3x)+(6x-1)(4x-10)=-50
\(\Rightarrow\)8x-24x2+2-6x+24x2-60x-4x+10=50
\(\Rightarrow\)-62x+12=50
\(\Rightarrow\)-62x=50-12
\(\Rightarrow\)-62x=38
\(\Rightarrow\)x=\(-\dfrac{38}{62}=-\dfrac{19}{31}\)
3)(x2+2x+4)(2-x)+x(x-3)(x+4)-x2+24=0
\(\Rightarrow\)8-x3+x(x2+4x-3x-12)-x2+24=0
\(\Rightarrow\)8-x3+x3+4x2-3x2-12x-x2+21=0
\(\Rightarrow\)-12x+29=0
\(\Rightarrow\)-12x=-29
\(\Rightarrow\)x=\(\dfrac{-29}{-12}=\dfrac{29}{12}\)
Tìm x
1. x2 - 5x + 6 = 0
2. (x + 4)2 - (3x - 1)2 = 0
3, x2 - 2x + 24 = 0
4, 9x2 - 4 = 0
5, x2 + 2x - 8 = 0
1.
\(x^2-5x+6=0\\ \Rightarrow x^2-2x-3x+6=0\\ \Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\\ \Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
2.
\(\left(x+4\right)^2-\left(3x-1\right)^2=0\\ \Rightarrow\left(x+4-3x+1\right)\left(x+4+3x-1\right)=0\\ \Rightarrow\left(-2x+5\right)\left(4x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}-2x+5=0\\4x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
3.
\(x^2-2x+24=0\\ \Rightarrow\left(x^2-2x+1\right)+23=0\\ \Rightarrow\left(x-1\right)^2+23=0\)
Vì (x-1)2≥0
23>0
\(\Rightarrow\left(x-1\right)^2+23>0\)
Vậy x vô nghiệm
4.
\(9x^2-4=0\\ \Rightarrow\left(3x-4\right)\left(3x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-4=0\\3x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)
5.
\(x^2+2x-8=0\\ \Rightarrow\left(x^2+2x+1\right)-9=0\\ \Rightarrow\left(x+1\right)^2-3^2=0\\ \Rightarrow\left(x-2\right)\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
a)9x2 – 49 = 0
b)(x – 1)(x + 2) – x – 2 = 0
c)(4x + 1)(x - 2) - (2x -3)(2x + 1) = 7
d)x(3x + 2) + (x + 1)2 – (2x – 5)(2x + 5) = 0
e)(x + 3)(x2 – 3x + 9) –x(x – 1)(x + 1) – 27 = 0
f)(4x-3)^2-3x(3-4x)=0
\(a,\Leftrightarrow\left(3x-7\right)\left(3x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\\ \Leftrightarrow-3x=6\Leftrightarrow x=-2\\ d,\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\\ \Leftrightarrow4x=-26\Leftrightarrow x=-\dfrac{13}{2}\\ e,\Leftrightarrow x^3+27-x^3+x-27=0\\ \Leftrightarrow x=0\\ f,\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
a) 9x2-49=0
(3x)2-72=0
<=> (3x-7)(3x+7)=0
th1: 3x-7=0
<=>3x=7
<=>x=\(\dfrac{7}{3}\)
th2: 3x+7=0
<=>3x=-7
<=>x=\(-\dfrac{7}{3}\)
giải phương trình sau
1/ 2x( x+3) - 6 (x-3) =0
2/ 2x^2( 2x+3) +(2x+3) =0
3/ (x-2) (x+1) -(x-2) 4x =0
4/ 2x ( x-5) -3x +15=0
5/ 3x(x+4) -2x-8 =0
6/ x^2 (2x-6) + 2x -6 =0
1: Ta có: \(2x\left(x+3\right)-6\left(x-3\right)=0\)
\(\Leftrightarrow2x^2+6x-6x+18=0\)
\(\Leftrightarrow2x^2+18=0\left(loại\right)\)
2: Ta có: \(2x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
hay \(x=-\dfrac{3}{2}\)
3: Ta có: \(\left(x-2\right)\left(x+1\right)-4x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(1-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
4: Ta có: \(2x\left(x-5\right)-3x+15=0\)
\(\Leftrightarrow\left(x-5\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
5: Ta có: \(3x\left(x+4\right)-2x-8=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
6: Ta có: \(x^2\left(2x-6\right)+2x-6=0\)
\(\Leftrightarrow2x-6=0\)
hay x=3
a) x2(x - 5) + 5 - x = 0; b) 3x4 - 9x3 = -9x2 + 27x;
c) x2(x + 8) + x2 = -8x; d) (x + 3)(x2 -3x + 5) = x2 + 3x.
e) 3x(x - 1) + x - 1 = 0;
f) (x - 2)(x2 + 2x + 7) + 2(x2 - 4) - 5(x - 2) = 0;
g) (2x - 1)2 - 25 = 0;
h) x3 + 27 + (x + 3)(x - 9) = 0.
i)8x3 - 50x = 0; k) 2(x + 3)-x2 - 3x = 0;
m)6x2 - 15x - (2x - 5)(2x + 5) =
a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
Giải các phương trình tích sau:
1.a)(3x – 2)(4x + 5) = 0 b) (2,3x – 6,9)(0,1x + 2) = 0
c)(4x + 2)(x2 + 1) = 0 d) (2x + 7)(x – 5)(5x + 1) = 0
2. a)(3x + 2)(x2 – 1) = (9x2 – 4)(x + 1)
b)x(x + 3)(x – 3) – (x + 2)(x2 – 2x + 4) = 0
c)2x(x – 3) + 5(x – 3) = 0 d)(3x – 1)(x2 + 2) = (3x – 1)(7x – 10)
3.a)(2x – 5)2 – (x + 2)2 = 0 b)(3x2 + 10x – 8)2 = (5x2 – 2x + 10)2
c)(x2 – 2x + 1) – 4 = 0 d)4x2 + 4x + 1 = x2
4. a) 3x2 + 2x – 1 = 0 b) x2 – 5x + 6 = 0
c) x2 – 3x + 2 = 0 d) 2x2 – 6x + 1 = 0
e) 4x2 – 12x + 5 = 0 f) 2x2 + 5x + 3 = 0
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
bài 2:
a, (3x+2)(x^2-1)=(9x^2-4)(x+1)
(3x+2)(x-1)(x+1)=(3x-2)(3x+2)(x+1)
(3x+2)(x-1)(x+1)-(3x-2)(3x+2)(x+1)=0
(3x+2)(x+1)(1-2x)=0
b, x(x+3)(x-3)-(x-2)(x^2-2x+4)=0
x(x^2-9)-(x^3+8)=0
x^3-9x-x^3-8=0
-9x-8=0
tự tìm x nha