cmr: (7x+1)2-(x+7)2=48(x2-1)
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
NA
2
NK
18 tháng 6 2017
(7x - 1)2 - (x+7)2 = 48(x2 - 1)
<=> 49x2 + 14x + 1 - x2 - 14x - 49 = 48x2 - 48
<=>48x2 - 48 = 48x2 - 48
<=> 0x = 0(luôn đúng)
Vậy ĐPCM
22 tháng 12 2022
a: \(\dfrac{2x^4-x^3-x^2+7x-4}{x^2+x-1}\)
\(=\dfrac{2x^4+2x^3-2x^2-3x^3-3x^2+3x+4x^2+4x-4}{x^2+x-1}\)
=2x^2-3x+4
b: \(=\dfrac{y}{x\left(2x-y\right)}+\dfrac{4x}{y\left(y-2x\right)}\)
\(=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(2x-y\right)\left(2x+y\right)}{xy\left(2x-y\right)}=\dfrac{-2x-y}{xy}\)
c: \(=\dfrac{6\left(x+8\right)}{7\left(x-1\right)}\cdot\dfrac{\left(x-1\right)^2}{\left(x-8\right)\left(x+8\right)}=\dfrac{6\left(x-1\right)}{7\left(x-8\right)}\)
KT
21 tháng 7 2018
BÀI 1:
Ta có: \(VT=\left(7x+1\right)^2-\left(x+7\right)^2\)
\(=\left(7x+1+x+7\right)\left(7x+1-x-7\right)\)
\(=\left(8x+8\right)\left(6x-6\right)\)
\(=8\left(x+1\right).6\left(x-1\right)\)
\(=48\left(x^2-1\right)=VP\) (đpcm)
Bài 2:
\(16x^2-\left(4x-5\right)^2=15\)
\(\Leftrightarrow\)\(16x^2-16x^2+40x-25=15\)
\(\Leftrightarrow\)\(40x=40\)
\(\Leftrightarrow\)\(x=1\)
Vậy...
Bài 3:
\(A=x^2+2x+3=\left(x+1\right)^2+2\ge2\)
Vậy MIN A = 2 khi x = -1
C
1
1 tháng 7 2021
a)4x2+4x+1-x2-10x-25=0
`<=>(2x+1)^2-(x+5)^2=0`
`<=>(2x+1-x-5)(2x+1+x+5)=0`
`<=>(x-4)(3x+6)=0`
`<=>(x-4)(x+2)=0`
`<=>` \(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\)
b)(x^2+x+7)(x^2+x-7)=(x2+x)2-7x
`<=>(x^2+x)^2-7^2=(x^2+x)^2-7x`
`<=>-7^2=-7x`
`<=>-49=-7x`
`<=>x=7`
Vậy x=7
TH
1
15 tháng 6 2018
Xét \(\left(7x+1\right)^2-\left(x+7\right)^2-48\left(x^2-1\right)\)
\(=49x^2+14x+1-x^2-14x-49-48x^2+48\)
\(=0\)
Vậy \(\left(7x+1\right)^2-\left(x+7\right)^2=48\left(x^2-1\right)\)
HN
1 tháng 7 2021
a)
⇔ \(x^2-16=9\)
⇔ \(x^2=25\)
⇔ \(x=\pm5\)
b)
⇔ \(x^2-4x+4-25x^2+20x-4=0\)
⇔ \(16x-24x^2=0\)
⇔ \(8x\left(2-3x\right)=0\)
⇒ \(\left[{}\begin{matrix}x=0\\2-3x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=\dfrac{2}{3}\)
c)
⇔ \(3x^2-10x-20=0\)
⇔ \(x^2-2.x.\dfrac{5}{3}+\dfrac{25}{9}-\dfrac{205}{9}=0\)
⇔ \(\left(x-\dfrac{5}{3}\right)^2=\dfrac{205}{9}\)
⇒ \(\left[{}\begin{matrix}x-\dfrac{5}{3}=\sqrt{\dfrac{205}{9}}\\x-\dfrac{5}{3}=-\sqrt{\dfrac{205}{9}}\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\\x=-\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\\\text{x}=-\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\end{matrix}\right.\)
Vậy...
d)
⇔ \(\left(x^2+x\right)^2-49=\left(x^2+x\right)^2-7x\)
⇔ 7x = 49
⇔ x=7
Vậy...
MT
1
18 tháng 7 2016
Theo đầu bài ta có:
\(\left(7x+1\right)^2-\left(x+7\right)^2=48\left(x^2-1\right)\)
\(\Rightarrow\left[\left(7x+1\right)+\left(x+7\right)\right]\left[\left(7x+1\right)-\left(x+7\right)\right]=\left(7^2-1^2\right)\left(x^2-1^2\right)\)
\(\Rightarrow\left(8x+8\right)\left(6x-6\right)=\left[\left(7+1\right)\left(7-1\right)\right]\left[\left(x+1\right)\left(x-1\right)\right]\)
\(\Rightarrow8\left(x+1\right)\cdot6\left(x-1\right)=8\left(x+1\right)\cdot6\left(x-1\right)\)( đpcm )
\(\left(7x+1\right)^2-\left(x+7\right)^2=\left(7x+1\right).\left(7x+1\right)-\left(x+7\right)\left(x+7\right)=\left(49x^2+7x+7x+1\right)-\left(x^2+7x+7x+49\right)\)\(=49x^2+14x+1-x^2-7x-7x-49=\left(49x^2-x^2\right)+\left(14x-7x-7x\right)-\left(49-1\right)=48x^2-48=48.\left(x^2-1\right)\)