chứng minh (a-b+c)2=a2+b2+c2-2ab-2bc+2ca
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=x^2-4x-5x+20\)
\(=x\left(x-4\right)-5\left(x-4\right)\)
\(=\left(x-4\right)\left(x-5\right)\)
x2 - 9x + 20
= x2 - 4x - 5x + 20
= x( x - 4 ) - 5( x - 4 )
= ( x - 4 )( x - 5 )
![](https://rs.olm.vn/images/avt/0.png?1311)
bỏ mũ 4
(x-2,5)+(x-1,5)=1
(x-5/2)+(x-3/2)=1
[x+(-5/2]+[x+(-3/2]=1
x^2+[(-5/2)+(-3/2)]=1
x^2+(-4)=1
X^2=1-(-4)
x^2=5
x^2=2,5^2
vậy x=2,5
![](https://rs.olm.vn/images/avt/0.png?1311)
( 4x - 1 )3 + ( 3 - 4x )( 9 + 12x + 16x2 ) = ( 8x - 1 )( 8x + 1 ) - ( 3x - 5 )
<=> 64x3 - 48x2 + 12x - 1 + [ 33 - ( 4x )3 ] = ( 8x )2 - 1 - 3x + 5
<=> 64x3 - 48x2 + 12x - 1 + 27 - 64x3 = 64x2 - 3x + 4
<=> -48x2 + 12x + 26 = 64x2 - 3x + 4
<=> -48x2 + 12x + 26 - 64x2 + 3x - 4 = 0
<=> -112x2 + 15x + 22 = 0 (*)
\(\Delta=b^2-4ac=15^2-4\cdot\left(-112\right)\cdot22=225+9856=10081\)
\(\Delta>0\)nên (*) có hai nghiệm phân biệt
\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{\sqrt{10081}-15}{-224}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-15-\sqrt{10081}}{-224}\end{cases}}\)
Lớp 8 sao nghiệm xấu thế -..-
![](https://rs.olm.vn/images/avt/0.png?1311)
(2x + 3)2 + (x - 1)(x + 1) = 5(x + 2)2 - (x - 5)(x + 1) + (x + 4)2
=> 4x2 + 12x + 9 + x2 - 1 = 5x2 + 20x + 20 - x2 + 4x + 5 + x2 + 8x + 16
=> 3x2 + 12x + 8 = 5x2 + 32x 41
=> -2x2 - 20x = 33
=> - 2x2 - 20x - 50 = -23
=> -2(x2 + 10x + 25) = -23
=> (x + 5)2 = 11,5
=> \(\orbr{\begin{cases}x=\sqrt{11,5}-5\\x=-\sqrt{11,5}-5\end{cases}}\)
\(4x^2+12x+9+x^2-1=5x^2+20x+20-\left(x^2-4x-5\right)+x^2+8x+16\)
\(5x^2+12x+8=6x^2+28x+36-x^2+4x+5\)
\(5x^2+12x+8=5x^2+32x+41\)
\(12x+8=32x+41\)
\(12x-32x=41-8\)
\(-20x=33\)
\(x=\frac{-33}{20}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm:
Đề này sai chắc rồi, đoạn cuối mạn phép sửa lại:
\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^n+1\right)+1\) ( với n là lũy thừa của 2 )
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^n+1\right)+1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^n+1\right)+1\)
Lm tg tự v.v...
\(=\left(2^n-1\right)\left(2^n+1\right)+1\)
\(=2^{2n}-1+1=2^{2n}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm:
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)
\(\Leftrightarrow a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+3abc-abc=0\)
\(\Leftrightarrow ab\left(a+b\right)+c\left(a^2+2ab+b^2\right)+c^2\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca+c^2\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
=> a+b=0 hoặc b+c=0 hoặc c+a=0
=> a=-b hoặc b=-c hoặc c=-a
=> Sẽ phải luôn tồn tại 2 trong 3 số a,b,c đối nhau
Ko mất tổng quát, g/s a=-b
=> \(\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=-\frac{1}{b^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{c^n}\) (vì n lẻ)
Và \(\frac{1}{\left(a+b+c\right)^n}=\frac{1}{\left(-b+b+c\right)^n}=\frac{1}{c^n}\)
=> \(\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{\left(a+b+c\right)^n}\)
( a - b + c )2
= [ ( a - b ) + c ]2
= ( a - b )2 + 2( a - b )c + c2
= a2 - 2ab + b2 + 2ac - 2bc + c2
= a2 + b2 + c2 - 2ab - 2bc + 2ca ( đpcm )
\(\left(a-b+c\right)^2\)
\(=\left(a-b+c\right).\left(a-b+c\right)\)
\(=a.\left(a-b+c\right)-b.\left(a-b+c\right)+c.\left(a-b+c\right)\)
\(=a^2-ab+ac-\left(ab-b^2+bc\right)+ac-bc+c^2\)
\(=a^2-ab+ac-ab+b^2-bc+ac-bc+c^2\)
\(=a^2-2ab+2ac+b^2-2bc+c^2\)
\(=a^2+b^2+c^2-2ab-2bc+2ac\)
\(\Rightarrow\left(a-b+c\right)^2=a^2+b^2+c^2-2ab-2bc+2ac\left(đpcm\right).\)