a) 3x2+12x-66=0
b) 9x2-30x+225=0
c) x2+3x-10=0
d) 3x2-7x+1=0
e) 3x2+7x+2=0
f) 4x2-12x+9=0
g) 3x2+7x+2=0
h) x2-4x+1=0
i) 2x2-6x+1=0
j) 3x2+4x-4=0
Cảm ơn bạn giải giúp mình rất nhiều .
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
Giải phương trình bằng cách đưa về phương trình tích :
3x2 + 2x - 1 = 0
x2 - 5x + 6 = 0
3x2 + 7x + 2 = 0
x2 - 4x + 1 = 0
2x2 - 6x + 1 = 0
3x2 + 4x - 4 = 0
3x2 + 2x - 1 = 0
=> 3x2 + 3x - x - 1 = 0
=> 3x(x + 1) - (x + 1) = 0
=> (3x - 1)(x + 1) = 0
=> \(\orbr{\begin{cases}3x-1=0\\x+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=-1\end{cases}}\)
x2 - 5x + 6 = 0
=> x2 - 2x - 3x + 6 = 0
=> x(x - 2) - 3(x - 2) = 0
=> (x - 3)(x - 2) = 0
=> \(\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=2\end{cases}}\)
3x2 + 7x + 2 = 0
=> 3x2 + 6x + x + 2 = 0
=> 3x(x + 2) + (x + 2) = 0
=> (3x + 1)(x + 2) = 0
=> \(\orbr{\begin{cases}3x+1=0\\x+2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{3}\\x=-2\end{cases}}\)
1, \(3x^2+2x-1=0\Leftrightarrow3x^2+3x-x-1=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\3x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}}\)
2, \(x^2-5x+6=0\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}}\)
3, \(3x^2+7x+2=0\Leftrightarrow3x^2+6x+x+2=0\)
\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\3x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{3}\end{cases}}}\)
\(x^2-4x+1=0\)
\(\Leftrightarrow\left(x^2-4x+4\right)=3\)
\(\Leftrightarrow\left(x-2\right)^2=3\)
\(\Leftrightarrow x=\sqrt{3}+2;x=2-\sqrt{3}\)
\(2x^2-6x+1=0\)
\(\Leftrightarrow4x^2-12x+2=0\)
\(\Leftrightarrow\left(2x-3\right)^2=7\)
\(\Leftrightarrow x=\frac{\sqrt{7}+3}{2};x=\frac{3-\sqrt{7}}{2}\)
\(3x^2+4x-4=0\)
\(\Leftrightarrow3x^2-2x+6x-4=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x-2\right)=0\)
\(\Leftrightarrow x=-2;x=\frac{2}{3}\)
giải phương trình sau:
a. (9x2-4)(x+1) = (3x+2) (x2-1)
b. (x-1)2-1+x2 = (1-x)(x+3)
c. (x2-1)(x+2)(x-3) = (x-1)(x2-4)(x+5)
d. x4+x3+x+1=0
e. x3-7x+6 = 0
f. x4-4x3+12x-9 = 0
g. x5-5x3+4x = 0
h. x4-4x3+3x2+4x-4 = 0
m.n jup vs
Giải các phương trình sau bằng máy tính bỏ túi (làm tròn kết quả đến chữ số thập phân thứ ba)
a) 2x2 - 5x - 4 = 0 ; b) -3x2 + 4x + 2 = 0
c) 3x2 + 7x + 4 = 0 ; d) 9x2 - 6x - 4 = 0.
Giải các phương trình tích sau
a) 3x2 + 12x – 66 = 0 b) 9x2 – 30x + 225 = 0
c) x2 + 3x – 10 = 0 d) 3x2 – 7x + 1 = 0
e) 3x2 – 7x + 8 = 0 f) 4x2 – 12x + 9 = 0
g) 3x2 + 7x + 2 = 0 h) x2 – 4x + 1 = 0
i) 2x2 – 6x + 1 = 0 j) 3x2 + 4x – 4 = 0
Giải các phương trình sau:
a) x − 1 = 3 x − 5 ;
b) x + 1 2 + 1 x + 3 = 0 ;
c) 3 x 2 − 4 x − 7 = 0 ;
d) 7 x − 1 2 x + 1 + 2 x + 1 x 2 − 1 = 0 .
a) 2x2 + 2x(5 - x)=12 d) 2(x + 5) - x2 - 5x = 0 g) (3x + 1)2 - (x+1) = 0
b) (5 - 2x)2 - 16 = 0 e) (2x - 1)2 - 4(x + 7)(x - 7) = 0 h) x2 + 7x - 8 = 0
c) 3x2 - 3x(x-2) = 36 f) (x + 4)2 - (x + 1)(x - 1) = 16 i) -2x2 +13x -15 = 0
mik cần gấp, cảm ơn mọi người.
\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)
\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)
1. x2+3x-8=0; 2). 8x2-2x-5=0 3). 2x2+7x+6=0 4). 3x2 -10x+8=0
5). 2x(8x-1)2(4x-1)=9
b. 4x2 +4x+1=0 d. 5x2 6x1=0 a. 2x2-5x+1=0 c. -3x2 +2x+8=0 e. -3x2+ 14x - 8=0 g. -7x2 +4x-3=0
a. 2x2-5x+1=0
△= b2 - 4ac = (-5)2 - 4*2*1 = 17 ⇒√△ = √17
\(\Rightarrow x_1=\frac{5+\sqrt{17}}{4};x_2=\frac{5-\sqrt{17}}{4}\)
Vậy .... S={\(\frac{5\pm\sqrt{17}}{4}\)}
b. 4x2 +4x+1=0
⇔(2x+1)2 = 0 ⇔ x=\(\frac{-1}{2}\)
c. -3x2 +2x+8=0
△' = b'2 - ac = 12 - (-3)*8 = 25 ⇒√△ = 5
\(\Rightarrow x_1=\frac{-1+5}{-3}=-\frac{4}{3};x_2=\frac{-1-5}{-3}=2\)
Vậy... S={-\(\frac{4}{3}\);2}
d. 5x2 6x1=0 (thiếu dấu nên mk chưa giải được)
e. -3x2+ 14x - 8=0
△' = b'2 - ac = 72 - (-3)*(-8) = 25 ⇒ √△ = 5
⇒\(x_1=\frac{-7+5}{-3}=\frac{2}{3};x_2=\frac{-7-5}{-3}=4\)
Vậy .... S={\(\frac{2}{3};4\)}
g. -7x2 +4x-3=0
△' = b'2 - ac = 22 - (-7)*(-3) = -17<0
Vậy pt vô nghiệm , S=∅