\(\dfrac{81}{16}\ge(\dfrac{3}{2})^n\ge\dfrac{4}{9}\)
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1: =>4a^3+4b^3-a^3-3a^2b-3ab^2-b^3>=0
=>a^3-a^2b-ab^2+b^3>=0
=>(a+b)(a^2-ab+b^2)-ab(a+b)>=0
=>(a+b)(a-b)^2>=0(luôn đúng)
2: \(a^4+b^4=\dfrac{a^4}{1}+\dfrac{b^4}{1}>=\dfrac{\left(a^2+b^2\right)^2}{1}=\dfrac{1}{2}\left(\dfrac{a^2}{1}+\dfrac{b^2}{1}\right)^2\)
=>\(a^4+b^4>=\dfrac{1}{2}\left(\dfrac{\left(a+b\right)^2}{2}\right)^2=\dfrac{\left(a+b\right)^4}{8}\)
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\(=\dfrac{5}{21}+\dfrac{16}{21}-\left(\dfrac{19}{23}+\dfrac{4}{23}\right)+\dfrac{1}{2}=\dfrac{1}{2}\)
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=>16x+9y=840 và 210/x-210/y=7/4
=>16x=840-9y và 30/x-30/y=1/4
=>x=-9/16y+52,5 và (30y-30x)=xy/4
=>xy=120y-120x
=>y(-9/16y+52,5)=120y-120(-9/16y+52,5)
=>-9/16y^2+52,5y-120y+120(-9/16y+52,5)=0
=>-9/16y^2-67,5y-67,5y+6300=0
=>y=40 hoặc y=-280
=>x=30 hoặc x=210
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=>9x+4y=360 và 36/x-36/y=1/2
=>4y=360-9x và 36/x-36/y=1/2
=>y=90-2,25x và \(\dfrac{36}{x}-\dfrac{36}{90-2,25x}=\dfrac{1}{2}\)
=>\(\dfrac{3240-81x-36x}{x\left(90-2,25x\right)}=\dfrac{1}{2}\)
=>90x-2,25x^2=2(3240-117x)
=>-2,25x^2+90x-6840+234x=0
=>x=118,3 hoặc x=25,7
=>y=-176,175 hoặc y=32,175
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Sửa đề: \(\dfrac{a^2+b^2}{2}\ge ab\)
Ta có: \(\left(a-b\right)^2\ge0\) với mọi a, b
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow a^2+b^2\ge2ab\)
\(\Leftrightarrow\dfrac{a^2+b^2}{2}\ge ab\)
Dấu "=" xảy ra khi a=b
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Với \(x\ge\dfrac{5}{2}\)có: \(A=x+\sqrt{2x-5}\ge\dfrac{5}{2}+0=\dfrac{5}{2}\)
Dấu '=' xảy ra \(\Leftrightarrow x=\dfrac{5}{2}\)
\(\Rightarrow A_{min}=\dfrac{5}{2}\)
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a) \(\dfrac{49}{81}=\dfrac{7^x}{9^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{7}{9}\right)^2=\left(\dfrac{7}{9}\right)^x\)\(\Rightarrow x=2\)
b) \(\dfrac{-64}{343}=\left(-\dfrac{4^x}{7^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{4}{7}\right)^3=\left(-\dfrac{4}{7}\right)^x\) \(\Rightarrow x=3\)
c) \(\dfrac{9}{144}=\dfrac{3^x}{12^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{3}{12}\right)^2=\left(\dfrac{3}{12}\right)^x\Rightarrow x=2\)
d) \(-\dfrac{1}{32}=\left(-\dfrac{1^x}{2^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^5=\left(-\dfrac{1}{2}\right)^x\Rightarrow x=5\)
Mong bạn xem lại đề bài.
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Trả lời:
\(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\\dfrac{2}{5}-3x=-\dfrac{3}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy x = -1/15; x = 1/3
`81/16 >= (3/2)^n >= 4/9`
`=> (3/2)^4 >= (3/2)^n >= (3/2)^2`
`=> 4 >= n >= 2`
`=> n = {2;3;4}`
Vậy: `n = {2;3;4}`