Em cần giúp bài nào thế? Mà em nên tách đăng 1 lần 1-2 bài thôi nha
Rút gọn :
a. ( x + 2 ) ( x2 - 2x + 4 ) - ( 1 - 3x ) ( 1 + 3x + 9x2)
b . ( x + y ) ( y2 - 2y + 4 ) + ( 5 - y ) ( 25 + 5y + y2)
a: =x^3+8-1+27x^3=28x^3+7
b: Sửa đề: (2+y)(y^2-2y+4)+(5-y)(25+5y+y^2)
=8+y^3+125-y^3
=133
Chứng minh đẳng thức (x-y)^2-4(x-y)(x+2y)+4(x+2y)^2
Cái bạn viết chưa phải đẳng thức. Bạn xem lại đề.
(3x-2)(3x+2)-9(x-1).x=0
\(\left(3x-2\right)\left(3x+2\right)-9\left(x+1\right)x=0\\ \Leftrightarrow\left[\left(3x\right)^2-2^2\right]-9x\left(x+1\right)=0\\ \Leftrightarrow\left(6x^2-4\right)\left(9x^2+9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}6x^2-4=0\\9x^2+9=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}6x^2=4\\9x^2=-9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=\dfrac{4}{6}\\x^2=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{2}{3}\\x^2=-1\left(v\text{ô}.l\text{ý}\right)\end{matrix}\right.\)
\(\left(3x-2\right)\left(3x+2\right)-9\left(x-1\right)x=0\)
\(\Leftrightarrow\left[\left(3x\right)^2-2^2\right]-9x\left(x-1\right)=0\)
\(\Leftrightarrow\left(9x^2-4\right)-9x^2+9x=0\)
\(\Leftrightarrow\left(9x^2-9x^2\right)+\left(9x-4\right)=0\)
\(\Leftrightarrow9x-4=0\)
\(\Leftrightarrow9x=4\)
\(\Leftrightarrow x=\dfrac{4}{9}\)
Vậy: \(x=\dfrac{4}{9}\)
tim gia tri nho nhat D= x^2 + 5y^2 -2xy + 4y +3
\(D=x^2+5y^2-2xy+4y+3\)
\(=x^2-2xy+y^2+4y^2+4y+1+2\)
\(=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+2\)
\(=\left(x-y\right)^2+\left(2y+1\right)^2+2\)
Vì \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\forall x,y\\\left(2y+1\right)^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-y\right)^2+\left(2y+1\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(2y+1\right)^2=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{1}{2}\)
Vậy \(D_{min}=2\Leftrightarrow x=y=-\dfrac{1}{2}\)
+) Cho a3 + b3 + c3 = 3abc. CMR: a + b + c = 0 và a = b = c
+) Áp dụng: Cho a3 + b3 + c3 = 3abc, vào bài toán:
Tính giá trị của biểu thức P= \(\dfrac{a+b}{c}\cdot\dfrac{b+c}{a}\cdot\dfrac{c+a}{b}\)
Bài 1:
$a^3+b^3+c^3=3abc$
$\Leftrightarrow (a+b)^3-3ab(a+b)+c^3-3abc=0$
$\Leftrightarrow [(a+b)^3+c^3]-[3ab(a+b)+3abc]=0$
$\Leftrightarrow (a+b+c)[(a+b)^2-c(a+b)+c^2]-3ab(a+b+c)=0$
$\Leftrightarrow (a+b+c)[(a+b)^2-c(a+b)+c^2-3ab]=0$
$\Leftrightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0$
$\Rightarrow a+b+c=0$ hoặc $a^2+b^2+c^2-ab-bc-ac=0$
Xét TH $a^2+b^2+c^2-ab-bc-ac=0$
$\Leftrightarrow 2(a^2+b^2+c^2)-2(ab+bc+ac)=0$
$\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0$
$\Rightarrow a-b=b-c=c-a=0$
$\Leftrightarrow a=b=c$
Vậy $a^3+b^3+c^3=3abc$ khi $a+b+c=0$ hoặc $a=b=c$
Áp dụng vào bài:
Nếu $a+b+c=0$
$A=\frac{-c}{c}+\frac{-b}{b}+\frac{-a}{a}=-1+(-1)+(-1)=-3$
Nếu $a=b=c$
$P=\frac{a+a}{a}+\frac{b+b}{b}+\frac{c+c}{c}=2+2+2=6$
Cho a+b+c=0. CMR: a3 + b3 + c3 = 3abc
rồi sau đó áp dụng: Tìm x, biết: (2x-2023)3 + (2020-x)3 + (23-x)3 = 0
a+b+c=0 nên a+b=-c
a^3+b^3+c^3
=(a+b)^3-3ab(a+b)+c^3
=(a+b+c)(a^2+2ab+b^2-bc-ac+c^2)-3ab(a+b)
=-3ab(-c)=3abc
(2x-2023)^3+(2020-x)^3+(23-x)^3=0
=>(2020-x)^3+(23-x)^3+[-(2020-x+23-x)^3]=0
=>3(2020-x)(23-x)(2x-2023)=0
=>\(x\in\left\{2020;23;\dfrac{2023}{2}\right\}\)
giúp mik với
a: A=(2-1)(2+1)(2^2+1)*...*(2^64+1)+1
=(2^2-1)(2^2+1)(2^4+1)*...*(2^64+1)+1
=(2^4-1)(2^4+1)*(2^8+1)*...(2^64+1)+1
=(2^8-1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)+1
...
=2^128-1+1=2^128
b: \(B=\dfrac{1}{2}\left[\left(3-1\right)\left(3+1\right)\cdot...\cdot\left(3^{64}+1\right)+2\right]\)
\(=\dfrac{1}{2}\left[\left(3^2-1\right)\left(3^2+1\right)\cdot...\cdot\left(3^{64}+1\right)+2\right]\)
\(=\dfrac{1}{2}\left[3^{128}-1+2\right]=\dfrac{3^{128}+1}{2}\)
c: \(=\left(a+b\right)^2+2c\left(a+b\right)+c^2+\left(a+b\right)^2-2c\left(a+b\right)+c^2-2\left(a+b\right)^2\)
=2c^2
Lời giải:
a,
$A=(100^2-99^2)+(98^2-97^2)+...+(2^2-1^2)$
$=(100-99)(100+99)+(98-97)(98+97)+...+(2-1)(2+1)$
$=100+99+98+97+...+2+1$
$=100(100+1):2=5050$
b.
$B=(2^2-1^2)+(4^2-3^2)+...+[n^2-(n-1)^2]$
$=(2-1)(2+1)+(4-3)(4+3)+...+[n-(n-1)][n+(n-1)]$
$=2+1+4+3+...+n+(n-1)$
$=1+2+3+....+(n-1)+n$
$=n(n+1):2$
https://hoc24.vn/cau-hoi/.8082461490144 (Bài 3)
Bài 1 :
\(a,\left(x-3\right)\left(x^2+3x+9\right)=x^3-3^3=x^3-27\)
\(b,\left(3x-1\right)\left(9x^2+3x+1\right)=\left(3x-1\right)\left[\left(3x\right)^2+3x.1+1\right]=\left(3x\right)^3-1^3=27x^3-1\)
\(c,\left(1-\dfrac{x}{2}\right)\left(1+\dfrac{x}{2}+\dfrac{x^2}{4}\right)\)
\(=\left(1-\dfrac{x}{2}\right)\left(1+\dfrac{x}{2}.1+\left(\dfrac{x}{2}\right)^2\right)\)
\(=1^3-\left(\dfrac{x}{2}\right)^3\)
\(=1-\dfrac{x^3}{8}\)
\(d,\left(\dfrac{x}{3}-y\right)\left(\dfrac{x^2}{9}+\dfrac{xy}{3}+y^2\right)\)
\(=\left(\dfrac{x}{3}-y\right)\left[\left(\dfrac{x}{3}\right)^2+\dfrac{x}{3}.y+y^2\right]\)
\(=\left(\dfrac{x}{3}\right)^3-y^3\)
\(=\dfrac{x^3}{27}-y^3\)