bài 5:
a) x2-10x+21=0
b)x2+8x+12=0
c) x2+9x+20=0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
TN
7
MH
29 tháng 3 2022
a) \(\text{Δ}=8^2-4.3.4=16\)
\(\left[{}\begin{matrix}x=\dfrac{-8+4}{2.3}=-\dfrac{2}{3}\\x=\dfrac{-8-4}{2.3}=-2\end{matrix}\right.\)
5 tháng 6 2021
`x^2+12=8x`
`<=>x^2-8x+12=0`
`<=>(x-2)(x-6)=0`
`<=>` $\left[ \begin{array}{l}x=2\\x=6\end{array} \right.$
`x^2-10x+12=0`
`<=>(x-5)^2-13=0`
`<=>` $\left[ \begin{array}{l}x=-\sqrt{13}+5\\x=\sqrt{13}+5\end{array} \right.$
Vậy không cso đáp án do đề sai
S
5 tháng 6 2021
Số nào dưới đây là nghiệm chung của hai phương trình \(x^2+12=8x\) và \(x^2-10x+12=0\) ?
Giải thích:
\(\left(1\right)x^2+12=8x\Leftrightarrow x^2-8x+12=0\)
\(\left(2\right)x^2-10x+12=0\)
Nghiệm của phương trình (1) là: \(\left\{{}\begin{matrix}x_1=6\\x_2=2\end{matrix}\right.\)
Nghiệm của phương trình (2) là: \(\left\{{}\begin{matrix}x_1=5+\sqrt{13}\\x_2=5-\sqrt{13}\end{matrix}\right.\)
\(\Rightarrow\) Không có nghiệm chung.
KV
9 tháng 12 2023
Bài 3
a) x² + 10x + 25
= x² + 2.x.5 + 5²
= (x + 5)²
b) 8x - 16 - x²
= -(x² - 8x + 16)
= -(x² - 2.x.4 + 4²)
= -(x - 4)²
c) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
d) (x + y)² - 9x²
= (x + y)² - (3x)²
= (x + y - 3x)(x + y + 3x)
= (y - 2x)(4x + y)
e) (x + 5)² - (2x - 1)²
= (x + 5 - 2x + 1)(x + 5 + 2x - 1)
= (6 - x)(3x + 4)
KV
9 tháng 12 2023
Bài 4
a) x² - 9 = 0
x² = 9
x = 3 hoặc x = -3
b) (x - 4)² - 36 = 0
(x - 4 - 6)(x - 4 + 6) = 0
(x - 10)(x + 2) = 0
x - 10 = 0 hoặc x + 2 = 0
*) x - 10 = 0
x = 10
*) x + 2 = 0
x = -2
Vậy x = -2; x = 10
c) x² - 10x = -25
x² - 10x + 25 = 0
(x - 5)² = 0
x - 5 = 0
x = 5
d) x² + 5x + 6 = 0
x² + 2x + 3x + 6 = 0
(x² + 2x) + (3x + 6) = 0
x(x + 2) + 3(x + 2) = 0
(x + 2)(x + 3) = 0
x + 2 = 0 hoặc x + 3 = 0
*) x + 2 = 0
x = -2
*) x + 3 = 0
x = -3
Vậy x = -3; x = -2
CM
25 tháng 8 2019
a) x = -1. b) x = 4 hoặc x = 5.
c) x = ± 2 . d) x = 1 hoặc x = 2.
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
18 tháng 10 2021
a/\(x^2+8x+17=\left(x^2+8x+16\right)+1=\left(x+4\right)^2+1\)
\(\left(x+4\right)^2\ge0\Rightarrow\left(x+4\right)^2+1\ge1>0\)
b/\(x^2-10x+29=\left(x^2-10x+25\right)+4=\left(x-5\right)^2+4\)
\(\left(x-5\right)^2\ge0\Rightarrow\left(x-5\right)^2+4\ge4>0\)
c/
11 tháng 4 2020
a/ \(3x(2x-3)=5(3-2x) \Leftrightarrow 3x(2x-3)+5(2x-3)=0 \\\ \Leftrightarrow (2x-3)(3x+5)=0 \)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=-\frac{5}{3}\end{matrix}\right.\)
KL: .............
b/ \(\left(x^2+1\right)\left(2x+5\right)=\left(x-1\right)\left(x^2+1\right)\Leftrightarrow\left(x^2+1\right)\left(2x+5\right)-\left(x-1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x+5-x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x+6=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=-6\end{matrix}\right.\)
KL: .............
c/ \(3x^3=x^2+3x-1\Leftrightarrow3x^3-x^2-3x+1=0\Leftrightarrow x^2\left(3x-1\right)-\left(3x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-1\right)=0\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=1\\x=-1\end{matrix}\right.\)
KL: ..........
d/ \(x^2-9x+20=0\Leftrightarrow x^2-5x-4x+20=0\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)
KL: .............
\(a.x^2-7x-3x+21=0\Leftrightarrow\left(x^2-7x\right)-\left(3x-21\right)=0\)
\(\Leftrightarrow x\left(x-7\right)-3\left(x-7\right)=0\Leftrightarrow\left(x-3\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=7\end{matrix}\right.\)
\(b.x^2+6x+2x+12=0\Leftrightarrow\left(x^2+6x\right)+\left(2x+12\right)=0\)
\(\Leftrightarrow x\left(x+6\right)+2\left(x+6\right)=0\Leftrightarrow\left(x+2\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-6\end{matrix}\right.\)
\(c.x^2+4x+5x+20=0\Leftrightarrow\left(x^2+4x\right)+\left(5x+20\right)=0\)
\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)=0\Leftrightarrow\left(x+5\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)