Bài 1:

Ta có: \(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3a^2b-3ab^2-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\left(do.a+b+c\ne0\right)\)

\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow a=b=c\)

\(M=\dfrac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\dfrac{3a^2}{\left(3a\right)^2}=\dfrac{3a^2}{9a^2}=\dfrac{1}{3}\)

Bài 1:

\(\left(a+b+c\right)^2=a^2+b^2+c^2\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)

\(\Leftrightarrow ab+bc+ac=0\Leftrightarrow bc=-ab-ac\)

\(\dfrac{a^2}{a^2+2bc}=\dfrac{a^2}{a^2+bc-ab-ac}=\dfrac{a^2}{\left(a-c\right)\left(a-b\right)}\)

CMTT: \(\left\{{}\begin{matrix}\dfrac{b^2}{b^2+2ca}=\dfrac{b^2}{\left(b-c\right)\left(b-a\right)}\\\dfrac{c^2}{c^2+2ab}=\dfrac{c^2}{\left(b-c\right)\left(a-c\right)}\end{matrix}\right.\)

\(M=\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=\dfrac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)

Bài 2:

\(a^3+b^3+c^3-3abc=\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3abc-3a^2b-3ab^2\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)(do \(a+b+c=0\))

\(\Rightarrow A=\dfrac{0}{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}=0\)

Là căn bậc 2 của:

a) \(2^2=4\)

b) \(\left(-7\right)^2=49\)

c) \(=\left(\sqrt{5}\right)^2=5\)

d) \(\left(\dfrac{5}{3}\right)^2=\dfrac{25}{9}\)

e) \(\sqrt{9}-8=3-8=-5\)

=> Là căn bậc 2 của: \(\left(-5\right)^2=25\)

g) \(\dfrac{1}{3}-\dfrac{2}{5}=-\dfrac{1}{15}\)

=> Là căn bậc 2 của: \(\left(-\dfrac{1}{15}\right)^2=\dfrac{1}{225}\)

a) \(=\left|-7\right|+\left|\dfrac{5}{4}\right|-\dfrac{3}{2}=7+\dfrac{5}{4}-\dfrac{3}{2}=\dfrac{27}{4}\)

b) \(=\dfrac{1}{2}.10-\dfrac{1}{4}+1=5-\dfrac{1}{4}+1=\dfrac{23}{4}\)

c) \(=\dfrac{2}{5}.\sqrt{\dfrac{1}{4}}-\sqrt{\dfrac{1}{4}}=\dfrac{2}{5}.\dfrac{1}{2}-\dfrac{1}{2}=\dfrac{1}{5}-\dfrac{1}{2}=-\dfrac{3}{10}\)

Lớp 4 chứ hc dấu "." đou =((

Tổng số kg ở 2 bao gạo: \(52,5\times2=105\left(kg\right)\)

Số kg gạo bao thứ nhất có: \(\left(105+3\right):2=54\left(kg\right)\)

Số kg gạo ở bao thứ 2: \(105-54=51\left(kg\right)\)

\(7110feet=7110.0,3048=2167,128\left(m\right)=2,167128\left(km\right)\)

\(d=80\sqrt{2h}=80\sqrt{2.2,167128}\approx166,6\left(km\right)\)

Câu 1:

a) \(=10x^3+15x^2-5x\)

b) \(=15x^2+14x-8-2x+2=15x^2+12x-6\)

Bài 2:

a) \(=5x\left(x-9\right)\)

b) \(=x\left(x-4\right)+y\left(x-4\right)=\left(x-4\right)\left(x+y\right)\)

c) \(=\left(x+2\right)^2-y^2=\left(x+2-y\right)\left(x+2+y\right)\)

Bài 3:

a) \(\Rightarrow\left(x-3\right)\left(5x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{5}\end{matrix}\right.\)

b) \(\Rightarrow9-x^2+x^2-2x+1=0\Rightarrow2x=10\Rightarrow x=5\)

\(\dfrac{R_1}{R_2}=\dfrac{S_2}{S_1}\Rightarrow R_2=\dfrac{R_1.S_1}{S_2}=\dfrac{17.10}{1}=170\left(\Omega\right)\)

Chọn B