\(\left\{{}\begin{matrix}a\left(a+b+c\right)=12\\b\left(a+b+c\right)=18\\c\left(a+b+c\right)=30\end{matrix}\right.\)

\(\Rightarrow a\left(a+b+c\right)+b\left(a+b+c\right)+c\left(a+b+c\right)=12+18+30\)

\(\Rightarrow\left(a+b+c\right)\left(a+b+c\right)=60\)

\(\Rightarrow\left(a+b+c\right)^2=60\)

\(\Rightarrow a+b+c=\pm\sqrt{60}\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=\sqrt{60}:12=\dfrac{\sqrt{15}}{6}\\b=\sqrt{60}:18=\dfrac{\sqrt{15}}{9}\\c=\sqrt{60}:30=\dfrac{\sqrt{15}}{15}\end{matrix}\right.\\\left\{{}\begin{matrix}a=-\sqrt{60}:12=\dfrac{-\sqrt{15}}{6}\\b=-\sqrt{60}:18=\dfrac{-\sqrt{15}}{9}\\c=-\sqrt{60}:30=\dfrac{-\sqrt{15}}{15}\end{matrix}\right.\end{matrix}\right.\)

Các câu sau làm tương tự

b. \(ab=\dfrac{3}{5};bc=\dfrac{4}{5};ac=\dfrac{3}{4}\)

\(\Rightarrow ab\cdot bc\cdot ac=\dfrac{9}{25}\Rightarrow\left(abc\right)^2=\dfrac{9}{25}\Rightarrow abc=\pm\dfrac{3}{5}\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=\dfrac{3}{5}:bc=\dfrac{3}{5}:\dfrac{4}{5}=\dfrac{3}{4}\\b=\dfrac{3}{5}:ac=\dfrac{3}{5}:\dfrac{3}{4}=\dfrac{4}{5}\\c=\dfrac{3}{5}:ab=\dfrac{3}{5}:\dfrac{3}{5}=1\end{matrix}\right.\\\left\{{}\begin{matrix}a=-\dfrac{3}{5}:\dfrac{4}{5}=-\dfrac{3}{4}\\b=-\dfrac{3}{5}:\dfrac{3}{4}=-\dfrac{4}{5}\\c=-\dfrac{3}{5}:\dfrac{3}{5}=-1\end{matrix}\right.\end{matrix}\right.\)

Vậy......................

3: Ta có \(\dfrac{1}{u_{n+1}}=\dfrac{1}{u_n}-1\).

Do đó \(\dfrac{1}{u_{100}}=\dfrac{1}{u_{99}}-1=\dfrac{1}{u_{98}}-2=...=\dfrac{1}{u_1}-99=\dfrac{1}{-2}-99=\dfrac{-199}{2}\Rightarrow u_{100}=\dfrac{-2}{199}\).

\(a^2=b^2+c^2-bc\Rightarrow bc=b^2+c^2-a^2\)

\(\Rightarrow cosA=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{bc}{2bc}=\dfrac{1}{2}\Rightarrow A=60^0\)

Tương tự: \(ac=a^2+c^2-b^2\Rightarrow cosB=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{1}{2}\Rightarrow B=60^0\)

\(\Rightarrow C=180^0-\left(A+B\right)=60^0\)

\(\Rightarrow A=B=C=60^0\Rightarrow\Delta ABC\) đều

Mới nghĩ ra 3 câu:

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

\(\sum\sqrt{\frac{ab}{1+c}}\le\sqrt{2\sum\frac{ab}{1+c}}\)

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

c/ \(ab+bc+ca=2abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

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

\(VT=\sum\frac{x^3}{\left(2-x\right)^2}\)

Ta có đánh giá: \(\frac{x^3}{\left(2-x\right)^2}\ge x-\frac{1}{2}\) \(\forall x\in\left(0;2\right)\)

\(\Leftrightarrow2x^3\ge\left(2x-1\right)\left(x^2-4x+4\right)\)

\(\Leftrightarrow9x^2-12x+4\ge0\Leftrightarrow\left(3x-2\right)^2\ge0\)

d/ Ta có đánh giá: \(\frac{x^4+y^4}{x^3+y^3}\ge\frac{x+y}{2}\)

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

Akai Haruma, Nguyễn Ngọc Lộc , @tth_new, @Băng Băng 2k6, @Trần Thanh Phương, @Nguyễn Việt Lâm

Mn giúp e vs ạ! Thanks!

Câu c quen thuộc, chém trước:

Ta có BĐT phụ: \(\frac{x^3}{x^3+\left(y+z\right)^3}\ge\frac{x^4}{\left(x^2+y^2+z^2\right)^2}\) \((\ast)\)

Hay là: \(\frac{1}{x^3+\left(y+z\right)^3}\ge\frac{x}{\left(x^2+y^2+z^2\right)^2}\)

Có: \(8(y^2+z^2) \Big[(x^2 +y^2 +z^2)^2 -x\left\{x^3 +(y+z)^3 \right\}\Big]\)

\(= \left( 4\,x{y}^{2}+4\,x{z}^{2}-{y}^{3}-3\,{y}^{2}z-3\,y{z}^{2}-{z}^{3 } \right) ^{2}+ \left( 7\,{y}^{4}+8\,{y}^{3}z+18\,{y}^{2}{z}^{2}+8\,{z }^{3}y+7\,{z}^{4} \right) \left( y-z \right) ^{2} \)

Từ đó BĐT \((\ast)\) là đúng. Do đó: \(\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\frac{x^2}{x^2+y^2+z^2}\)

\(\therefore VT=\sum\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\sum\frac{x^2}{x^2+y^2+z^2}=1\)

Done.

Câu 1 chuyên phan bội châu

câu c hà nội

câu g khoa học tự nhiên

câu b am-gm dựa vào hằng đẳng thử rồi đặt ẩn phụ

câu f đặt \(a=\frac{2m}{n+p};b=\frac{2n}{p+m};c=\frac{2p}{m+n}\)

Gà như mình mấy câu còn lại ko bt nha ! để bạn tth_pro full cho nhé !

Nguyễn Ngọc Lộc , ?Amanda?, Phạm Lan Hương, Akai Haruma, @Trần Thanh Phương, @Nguyễn Việt Lâm,

@tth_new

Giúp em vs ạ! Thanks nhiều ạ

Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)

\(\Leftrightarrow a^2b+ab^2+c^2a+ca^2+b^2c+bc^2+2abc=0\)

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

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

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

=> Hoặc a+b=0 hoặc b+c=0 hoặc c+a=0

=> Hoặc a=-b hoặc b=-c hoặc c=-a

Ko mất tổng quát, g/s a=-b

a) Ta có: vì a=-b thay vào ta được:

\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{1}{b^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{c^3}\)

\(\frac{1}{a^3+b^3+c^3}=\frac{1}{-b^3+b^3+c^3}=\frac{1}{c^3}\)

=> đpcm

b) Ta có: \(a+b+c=1\Leftrightarrow-b+b+c=1\Rightarrow c=1\)

=> \(P=-\frac{1}{b^{2021}}+\frac{1}{b^{2021}}+\frac{1}{c^{2021}}=\frac{1}{1^{2021}}=1\)

Lời giải:
a)

Đặt \(\frac{x-12}{4}=\frac{y-9}{3}=z-1=k\Rightarrow \left\{\begin{matrix} x=4k+12\\ y=3k+9\\ z=k+1\end{matrix}\right.\)

Khi đó:

\(3x+5y-z=2\)

\(\Leftrightarrow 3(4k+12)+5(3k+9)-(k+1)=2\)

$\Rightarrow k=-3$

\(\Rightarrow \left\{\begin{matrix} x=4k+12=0\\ y=3k+9=0\\ z=k+1=-2\end{matrix}\right.\)

b)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{a+b}{6}=\frac{b+c}{7}=\frac{a+c}{8}=\frac{a+b+b+c+c+a}{6+7+8}=\frac{2(a+b+c)}{21}=\frac{2.14}{21}=\frac{4}{3}\)

\(\Rightarrow \left\{\begin{matrix} a+b=8\\ b+c=\frac{28}{3}\\ c+a=\frac{32}{3}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} a+b=8\\ b+c=\frac{28}{3}\\ c+a=\frac{32}{3}\\ a+b+c=14\end{matrix}\right.\Rightarrow \left\{\begin{matrix} c=6\\ a=\frac{14}{3}\\ b=\frac{10}{3}\end{matrix}\right.\)

Lời giải:
a)

Đặt \(\frac{x-12}{4}=\frac{y-9}{3}=z-1=k\Rightarrow \left\{\begin{matrix} x=4k+12\\ y=3k+9\\ z=k+1\end{matrix}\right.\)

Khi đó:

\(3x+5y-z=2\)

\(\Leftrightarrow 3(4k+12)+5(3k+9)-(k+1)=2\)

$\Rightarrow k=-3$

\(\Rightarrow \left\{\begin{matrix} x=4k+12=0\\ y=3k+9=0\\ z=k+1=-2\end{matrix}\right.\)

b)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{a+b}{6}=\frac{b+c}{7}=\frac{a+c}{8}=\frac{a+b+b+c+c+a}{6+7+8}=\frac{2(a+b+c)}{21}=\frac{2.14}{21}=\frac{4}{3}\)

\(\Rightarrow \left\{\begin{matrix} a+b=8\\ b+c=\frac{28}{3}\\ c+a=\frac{32}{3}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} a+b=8\\ b+c=\frac{28}{3}\\ c+a=\frac{32}{3}\\ a+b+c=14\end{matrix}\right.\Rightarrow \left\{\begin{matrix} c=6\\ a=\frac{14}{3}\\ b=\frac{10}{3}\end{matrix}\right.\)

a.

\(\frac{3x-36}{12}=\frac{5y-45}{15}=\frac{z-1}{1}=\frac{3x+5y-z-50}{26}=\frac{-48}{26}\)

\(\Rightarrow\frac{x-12}{4}=\frac{-48}{26}\Rightarrow x=...\)

Tương tự với y, z, nhưng chắc bạn nhầm đề, nếu pt bên dưới là -2 thì nó ra \(\frac{-52}{26}=-2\) kết quả đẹp hơn nhiều

b. Không rõ đề

c.

\(x+y+z=9\Rightarrow\left(x+y+z\right)^2=81=3.27=3\left(xy+yz+zx\right)\)

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

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)

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

\(\Leftrightarrow x=y=z\Rightarrow\frac{3}{x}=1\Rightarrow x=y=z=3\)