Choose the letter A, B, C or D the word that has the underlined part different from others.

A. homework

B. go 

C. sports

D. told

Câu a: tự chứng minh.

Câu b: áp dụng câu a

thay xyz=(4-x-y-z)²vào

có: \(\dfrac{1}{c}=\dfrac{2}{b}-\dfrac{1}{a}=\dfrac{2a-b}{ab}\Rightarrow2a-b=\dfrac{ab}{c}\)

tương tự ta cũng có \(2c-b=\dfrac{bc}{a}\)

\(VT=\dfrac{c\left(a+b\right)}{ab}+\dfrac{a\left(c+b\right)}{bc}=\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}+\dfrac{a}{c}=\left(\dfrac{c}{a}+\dfrac{a}{c}\right)+\dfrac{a+c}{b}\)

Áp dụng BĐt AM-GM:\(\dfrac{c}{a}+\dfrac{a}{c}\ge2\)

và \(\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}\ge\dfrac{4}{a+c}\Leftrightarrow a+c\ge2b\)

do đó \(VT\ge2+2=4\)

Dấu = xảy ra khi a=b=c

ta có :\(a^2-ab+b^2=\left(a+b\right)^2-3ab\ge\left(a+b\right)^2-\dfrac{3}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)(theo BĐT AM-GM)

\(\Rightarrow P\ge\sum\dfrac{a+b}{2\sqrt{ab+1}}\)

ÁP dụng BĐT AM-GM:

\(\dfrac{a+b}{2\sqrt{ab+1}}+\dfrac{b+c}{2\sqrt{bc+1}}+\dfrac{c+a}{2\sqrt{ca+1}}\ge3\sqrt[3]{\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{8\sqrt{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}}}=\dfrac{3}{2}.\dfrac{1}{\sqrt[3]{\sqrt{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}}}\)

Mà \(\sqrt[3]{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}\le\dfrac{1}{3}\left(ab+bc+ca+3\right)\)

\(\Rightarrow P\ge\dfrac{3\sqrt{3}}{2\sqrt{\left(ab+bc+ca+3\right)}}\)(*)

ta liên tưởng đến BĐT phụ:\(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\)

Cm: phân tích :\(VT=xy\left(x+y\right)+yz\left(y+z\right)+zx\left(x+z\right)+2xyz\)

\(=xy\left(x+y\right)+yz\left(y+z\right)+xz\left(z+x\right)+3xyz-xyz\)

\(=\left(x+y+z\right)\left(xy+yz+xz\right)-xyz\)

mà \(\left(x+y+z\right)\left(xy+yz+xz\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{x^2y^2z^2}=9xyz\)

nên \(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\left(x+y+z\right)\left(xy+yz+xz\right)-\dfrac{1}{9}\left(x+y+z\right)\left(xy+yz+xz\right)=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)

Áp dụng:

\(1=\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\dfrac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)

mặt khác,theo AM-GM,dễ dàng chứng minh được \(a+b+c\ge\dfrac{3}{2}\)

nên \(1\ge\dfrac{8}{9}.\dfrac{3}{2}\left(ab+bc+ca\right)\Leftrightarrow ab+bc+ca\le\dfrac{3}{4}\)

từ (*)\(\Rightarrow P\ge\dfrac{3\sqrt{3}}{2\sqrt{\dfrac{3}{4}+3}}=\dfrac{3}{\sqrt{5}}\)

Dấu = xảy ra khi \(a=b=c=\dfrac{1}{2}\)

\(PT\Leftrightarrow3^x=1+y^3=\left(y+1\right)\left(y^2-y+1\right)\)

Nhận thấy VT là số lẻ \(\rightarrow\)VF phải là số lẻ ,suy ra y chẵn nên \(y\ge2\)

y chẵn, đặt \(y=2y_1\left(y_1\in N\right)\),Pt đầu trở thành \(3^x-8y_1^3=1\)

\(\Rightarrow8y_1^3⋮4\)\(\Rightarrow3^x\equiv1\left(mod4\right)\)nên x chẵn . Đặt \(x=2x_1\left(x_1\in N\right)\)

\(PT\Leftrightarrow\left(3^{x_1}\right)^2-1=y^3\Leftrightarrow\left(3^{x_1}-1\right)\left(3^{x_1}+1\right)=y^3\)

tồn tại m,n sao cho \(3^{x_1}-1=y^m;3^{x_1}+1=y^n\left(m,n\in N;m< n;m+n=3\right)\)

\(\Rightarrow y^n-y^m=2=y^m\left(y^{n-m}-1\right)\)

với y>2 thì \(y^m\left(y^{n-m}-1\right)>2\)nên điều này chỉ xảy ra khi y=2

thay vào: \(2^m\left(2^{n-m}-1\right)=2\)\(\Leftrightarrow\left\{{}\begin{matrix}m=1\\n=2\end{matrix}\right.\)( m+n=3)

do đó x=2

Vậy (x;y)=(2;2)

P/s: không chắc chắn

a) =48

b) =0

Tick mình nhé,mình trả lời chính xác lắm

\(\left\{{}\begin{matrix}x^2+y^2+z^2=1\left(1\right)\\x^2+y^2-2xy+2yz-2xz+1=0\left(2\right)\end{matrix}\right.\)

Thay \(1=x^2+y^2+z^2\)vào phương trình (2):

\(2x^2+2y^2+z^2-2xy+2yz-2xz=0\)

\(\Leftrightarrow\left(x-y-z\right)^2+x^2+y^2=0\)

mà \(\left(x-y-z\right)^2;x^2;y^2\)không âm

\(\Rightarrow\left\{{}\begin{matrix}x-y-z=0\\x=0\\y=0\end{matrix}\right.\)\(\Leftrightarrow x=y=z=0\)(mâu thuẫn với (1))

Vậy HPT vô nghiệm