\(\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+b^2\ge5\sqrt[5]{\dfrac{a^{20}b^2}{b^{12}}}=5.\dfrac{a^4}{b^2}\)

\(\Rightarrow4.\dfrac{a^5}{b^3}+b^2\ge5.\dfrac{a^4}{b^2}\)

Tương tự: \(4.\dfrac{b^5}{c^3}+c^2\ge5\dfrac{b^4}{c^2};4\dfrac{c^5}{a^3}+a^2\ge5.\dfrac{c^4}{a^2}\)

\(\Rightarrow4\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+a^2+b^2+c^2\ge5\left(\dfrac{c^4}{a^2}+\dfrac{a^4}{b^2}+\dfrac{b^4}{c^2}\right)\)

Lại có: \(\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+b^2+b^2+b^2\ge5a^2\)

\(\Rightarrow2.\dfrac{a^5}{b^3}+3b^2\ge5a^2\), tương tự: \(2.\dfrac{b^5}{c^3}+3c^2\ge5b^2;2\dfrac{c^5}{a^3}+3a^2\ge5c^2\)

\(\Rightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\ge a^2+b^2+c^2\)

\(\Rightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}+4.\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge4.\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+a^2+b^2+c^2\ge5.\left(\dfrac{c^4}{a^2}+\dfrac{a^4}{b^2}+\dfrac{b^4}{c^2}\right)\)

\(\Rightarrow dpcm\)

giả sử \(a>b>c>0\) thì ta có :

\(\dfrac{a^4}{b^2}\left(\dfrac{a}{b}-1\right)+\dfrac{b^4}{c^2}\left(\dfrac{b}{c}-1\right)+\dfrac{c^4}{a^2}\left(\dfrac{c}{a}-1\right)\ge\dfrac{2a^2b}{c}+\dfrac{c^5}{a^3}-\dfrac{c^4}{a^2}\)

\(\ge\dfrac{2c^4b}{a}-\dfrac{c^4}{a^2}=\dfrac{c^4}{a}\left(2b-\dfrac{1}{a}\right)>0\)

làm tương tự cho trường hợp \(c>b>a>0\) ; \(b>a>c\) và \(b>c>a\)

\(\Rightarrow\left(đpcm\right)\)

mấy câu cậu câu đăng khác bn làm tương tự nha . nếu bn lm không được thì có j mk lm luôn cho còn h mk bạn rồi :(

2:

a: Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{24}{9}=\dfrac{8}{3}\)

=>x=16/3; y=8; z=32/3

A=3x+2y-6z

=3*16/3+2*8-6*32/3

=16+16-64

=-32

b: Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y+z}{5-6+7}=\dfrac{6\sqrt{2}}{6}=\sqrt{2}\)

=>x=5căn 2; y=6căn 2; y=7căn 2

B=xy-yz

=y(x-z)

=6căn 2(5căn 2-7căn 2)

=-6căn 2*2căn 2

=-24

bài 1 a)áp dụng dãy tỉ số bằng nhau ta có:\(\dfrac{a+b+c}{3+4+5}\)=\(\dfrac{24}{12}\)=2

a=2.3=6 ; b=2.4=8 ;c=2.5=10

M=ab+bc+ac=6.8+8.10+6.10=48+80+60=188

"nhưng bài còn lại làm tương tự"

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=b.k\\c=d.k\end{matrix}\right.\)

Ta có:

\(\dfrac{4.a-5.b}{4.a+5.b}=\dfrac{4.a+5.b-10.b}{4.a+5.b}=1-\dfrac{10.b}{4.a+5.b}=1-\dfrac{10.b}{4.b.k+5b}=1-\dfrac{10}{4.k+5}\) (1)

\(\dfrac{4.c-5.d}{4.c+5.d}=\dfrac{4.c+5.d-10.d}{4.c+5.d}=1-\dfrac{10.d}{4.c+5.d}=1-\dfrac{10.d}{4.d.k+5.d}=1-\dfrac{10}{4.k+5}\) (2)

Từ (1) và (2) suy ra \(\dfrac{4.a-5.b}{4.a+5.b}=\dfrac{4.c-5.d}{4.c+5.d}\left(đpcm\right)\)

Lời giải:

Đặt \(\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt\)

Khi đó ta có:

\(\frac{4a-5b}{4a+5b}=\frac{4bt-5b}{4bt+5b}=\frac{b(4t-5)}{b(4t+5)}=\frac{4t-5}{4t+5}\)

\(\frac{4c-5d}{4c+5d}=\frac{4dt-5d}{4dt+5d}=\frac{d(4t-5)}{d(4t+5)}=\frac{4t-5}{4t+5}\)

Do đó: \(\frac{4a-5b}{4a+5b}=\frac{4c-5d}{4c+5d}\) (đpcm)

a) = =

b) = = = . ( Với điều kiện b # 1)

c) \(\dfrac{a^{\dfrac{1}{3}}b^{-\dfrac{1}{3}-}a^{-\dfrac{1}{3}}b^{\dfrac{1}{3}}}{\sqrt[3]{a^2}-\sqrt[3]{b^2}}\)= = = ( với điều kiện a#b).

d) \(\dfrac{a^{\dfrac{1}{3}}\sqrt{b}+b^{\dfrac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}\) = = = =

a) \(A=\dfrac{7}{8}\left(-\dfrac{5}{9}-\dfrac{4}{9}\right)+5\dfrac{7}{8}\)

\(A=\dfrac{7}{8}.\left(-1\right)+5\dfrac{7}{8}=5\dfrac{7}{8}-\dfrac{7}{8}=5\).

\(B=\dfrac{1}{4}.\dfrac{8}{5}.\dfrac{25}{16}.\dfrac{-7}{4}=\dfrac{-35}{32}\)

a) \(A=\dfrac{-7}{8}.\dfrac{5}{9}-\dfrac{4}{9}.\dfrac{7}{8}+5\dfrac{7}{8}\) 

\(A=\dfrac{7}{8}.\left(\dfrac{-5}{9}-\dfrac{4}{9}\right)+\dfrac{47}{8}\) 

\(A=\dfrac{7}{8}.-1+\dfrac{47}{8}\) 

\(A=\dfrac{-7}{8}+\dfrac{47}{8}\) 

\(A=5\) 

b) \(B=0,25.1\dfrac{3}{5}.\left(\dfrac{5}{4}\right)^2:\left(\dfrac{-4}{7}\right)\) 

\(B=\dfrac{1}{4}.\dfrac{8}{5}.\dfrac{25}{16}.\dfrac{-7}{4}\) 

\(B=\dfrac{-35}{32}\)