Bạn chỉ cần bình phương PT x/a + y/b + z/c 

và chỉ ra ayz + bxz + cxy = 0 ở PT 2 là xong 

:D 

\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow ayz+bxz+cxy=0\)

\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Rightarrow(\frac{x}{a}+\frac{y}{b}+\frac{z}{c})^2=1\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac})=1\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1-2(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac})=1-2\frac{ayz+bxz+cxy}{abc}=1-2\cdot0=1(đpcm)\)

Câu hỏi của Tăng Thiện Đạt - Toán lớp 8 - Học toán với OnlineMath

Tham khảo nhé mk làm rồi !

\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow ayz+bxz+cxy=0\)

\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Leftrightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)

\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1-2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)\)

\(=1-2.\frac{cxy+bxz+ayz}{abc}=1-2.0=1\)

khó quá ta

Đặt : x/a = m ; y/b = n ; z/c = p

=> m+n+p = 1 ; 1/m+1/n+1/p=0

1/m+1/n+1/p=0

<=> mn+np+pm/mnp=0

<=> mn+np+pm=0

<=> 2mn+2np+2pm=0

Xét : 1 = (m+n+p)^2 = m^2+n^2+p^2+2mn+2np+2pm = m^2+n^2+p^2

=> x^2/a^2+y^2/b^2+z^2/c^2 = 1

=> ĐPCM

Tk mk nha

Ta có: \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)

\(\Rightarrow\frac{ayz}{xyz}+\frac{bxz}{xyz}+\frac{cxy}{xyz}=0\)

\(\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\)

\(\Rightarrow ayz+bxz+cxy=0\left(1\right)\)

Mặt khác: \(\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{cxy+ayz+bxz}{abc}\right)=1\left(2\right)\)

Thay (1) vào (2) \(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2.\frac{0}{abc}=1\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(đpcm\right)\)

Ta có 

\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)

\(\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\)

\(\Rightarrow ayz+bxz+cxy=0\)

Ta có

\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)

\(\Rightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+\frac{2xy}{ab}+\frac{2yz}{bc}+\frac{2xz}{ac}=1\)

\(\Rightarrow\frac{2xy}{ab}+\frac{2yz}{bc}+\frac{2xz}{ac}=1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)

\(\Rightarrow\frac{2xy.abc^2+2yz.a^2bc+2xz.ab^2c}{a^2b^2c^2}=1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)

\(\Rightarrow\frac{2abc.\left(cxy+ayz+bxz\right)}{a^2b^2c^2}=1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)

 Ta có \(cxy+ayz+bxz=0\)

\(\Rightarrow\frac{2abc.\left(cxy+ayz+bxz\right)}{a^2b^2c^2}=1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)

\(\Rightarrow\frac{2abc.0}{a^2b^2c^2}=1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)\)

\(\Rightarrow1-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)=0\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(đpcm\right)\)

bài này bạn bình phương vế thứ 2 lên rồi phân k vế 1 là ra đấy

Vì đã khuya nên não cũng không còn hoạt động tốt nữa, mình làm bài 1 thôi nhé.

Bài 1:

a)

\(2\text{VT}=\sum \frac{2bc}{a^2+2bc}=\sum (1-\frac{a^2}{a^2+2bc})=3-\sum \frac{a^2}{a^2+2bc}\)

Áp dụng BĐT Cauchy-Schwarz:

\(\sum \frac{a^2}{a^2+2bc}\geq \frac{(a+b+c)^2}{a^2+2bc+b^2+2ac+c^2+2ab}=\frac{(a+b+c)^2}{(a+b+c)^2}=1\)

Do đó: \(2\text{VT}\leq 3-1\Rightarrow \text{VT}\leq 1\) (đpcm)

Dấu "=" xảy ra khi $a=b=c$

b)

Áp dụng BĐT Cauchy-Schwarz:

\(\text{VT}=\sum \frac{ab^2}{a^2+2b^2+c^2}=\sum \frac{ab^2}{\frac{a^2+b^2+c^2}{3}+\frac{a^2+b^2+c^2}{3}+\frac{a^2+b^2+c^2}{3}+b^2}\leq \sum \frac{1}{16}\left(\frac{9ab^2}{a^2+b^2+c^2}+\frac{ab^2}{b^2}\right)\)

\(=\frac{1}{16}.\frac{9(ab^2+bc^2+ca^2)}{a^2+b^2+c^2}+\frac{a+b+c}{16}(1)\)

Áp dụng BĐT AM-GM:

\(3(ab^2+bc^2+ca^2)\leq (a^2+b^2+c^2)(a+b+c)\)

\(\Rightarrow \frac{1}{16}.\frac{9(ab^2+bc^2+ca^2)}{a^2+b^2+c^2)}\leq \frac{3}{16}(a+b+c)(2)\)

Từ $(1);(2)\Rightarrow \text{VT}\leq \frac{a+b+c}{4}$ (đpcm)

Dấu "=" xảy ra khi $a=b=c$

Bài 2/Áp dụng BĐT Bunyakovski:

\(\left(x^2+y^2+z^2\right)\left(1^2+3^2+5^2\right)\ge\left(x+3y+5z\right)^2\)

\(\Rightarrow P\ge\frac{\left(x+3y+5z\right)^2}{35}\) (*)

Ta có: \(x+3y+5z=x.1+\frac{y}{3}.9+\frac{z}{5}.25\)

\(=\frac{16z}{5}+8\left(\frac{y}{3}+\frac{z}{5}\right)+1\left(\frac{z}{5}+\frac{y}{3}+x\right)\)

\(\ge16+8.2+1.3=35\). Thay vào (*) là xong.

Đẳng thức xảy ra khi x = 1; y =3; z = 5

No choice teen, Akai Haruma, Arakawa Whiter, Phạm Lan Hương, soyeon_Tiểubàng giải, tth, Nguyễn Văn Đạt

@Nguyễn Việt Lâm

giúp em với ạ! Cần gấp lắm! Thanks nhiều!

bài này chúa Pain làm rất nhiều lần rồi ? m ko biết ấn vào câu hỏi tương tự để xem ak 

https://olm.vn/hoi-dap/question/1159233.html.

\(\frac{a^2}{x}+\frac{b^2}{y}=\frac{\left(a+b\right)^2}{x+y}\) " C/M 2 số rồi suy ra 3 số cx như vậy "

\(\frac{a^2x+b^2y}{xy}=\frac{\left(a+b\right)^2}{x+y}\) " Quy đồng VT "

\(\left(a^2x+b^2y\right)\left(x+y\right)=xy\left(a+b\right)\left(a+b\right)\) " nhân chéo mẫu số "

\(a^2x^2+a^2xy+b^2y^2+b^2xy=a^2xy+2abxy+b^2xy.\)

\(\left(a^2x^2-2abxy+b^2y^2\right)+\left(a^2xy-a^2xy\right)+\left(b^2xy-b^2xy\right)=0\)

\(\left(ax-by\right)^2=0\) " đúng " dcpcm

a/ \(\Leftrightarrow\frac{x+y}{xy}\ge\frac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow x^2+y^2-2xy\ge0\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)

Vậy BĐT đã cho đúng

b/ \(\frac{a}{a+b^2}=\frac{a}{a\left(a+b+c\right)+b^2}=\frac{a}{a^2+b^2+a\left(b+c\right)}\le\frac{a}{2ab+a\left(b+c\right)}=\frac{1}{b+b+b+c}\)

\(\Rightarrow\frac{a}{a+b^2}=\frac{1}{b+b+b+c}\le\frac{1}{16}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{16}\left(\frac{3}{b}+\frac{1}{c}\right)\)

Tương tự: \(\frac{b}{b+c^2}\le\frac{1}{16}\left(\frac{3}{c}+\frac{1}{a}\right)\) ; \(\frac{c}{c+a^2}\le\frac{1}{16}\left(\frac{3}{a}+\frac{1}{c}\right)\)

Cộng vế với vế:

\(VT\le\frac{1}{16}\left(\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\right)=\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)

câu 1 là :từ a/x + b/y + c/z =0 suy ra (ayz+bxz+cxy)/xyz =0 suy ra ayz+bxz+cxy=0 (1)

vì x/a + y/b + z/c =1 (gt) suy ra (x/a + y/b + z/c )^2 = 1^2 . suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 + 2(xy/ab + yz/bc + xz/ac) =1

suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 + 2[(ayz+bxz+cxy)/abc = 1 (2)

Từ (1) và (2) suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 =1 (đpcm)

câu 3 98

Tks pạn nhìu @!

Ta có:   \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\)

\(\Leftrightarrow\)\(\frac{bcx+acy+abz}{abc}=0\)

\(\Leftrightarrow\)\(bcx+acy+abz=0\)

           \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\)

\(\Leftrightarrow\)\(\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)^2=4\)

\(\Leftrightarrow\)\(\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}+2\left(\frac{ab}{xy}+\frac{ac}{xz}+\frac{bc}{yz}\right)=4\)

\(\Leftrightarrow\)\(\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}=4-2\frac{abz+acy+bcx}{xyz}=4\)                (vì  abz + acy + bcx = 0 )

1.a>0.√a

2.c/mb/z+x/y=a/b6

=x/y=y/x

4.xxy/2 2

5.a/b+ab=ab2