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Câu 1:
\(4\sqrt[4]{\left(a+1\right)\left(b+4\right)\left(c-2\right)\left(d-3\right)}\le a+1+b+4+c-2+d-3=a+b+c+d\)
Dấu = xảy ra khi a = -1; b = -4; c = 2; d= 3
\(\frac{a^2}{b^5}+\frac{1}{a^2b}\ge\frac{2}{b^3}\)\(\Leftrightarrow\)\(\frac{a^2}{b^5}\ge\frac{2}{b^3}-\frac{1}{a^2b}\)
\(\frac{2}{a^3}+\frac{1}{b^3}\ge\frac{3}{a^2b}\)\(\Leftrightarrow\)\(\frac{1}{a^2b}\le\frac{2}{3a^3}+\frac{1}{3b^3}\)
\(\Rightarrow\)\(\Sigma\frac{a^2}{b^5}\ge\Sigma\left(\frac{5}{3b^3}-\frac{2}{3a^3}\right)=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\)
\(\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}=\frac{a^4}{ab+ac}+\frac{b^4}{ab+bc}+\frac{c^4}{ac+bc}\)
\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}\)
\(=\frac{a^2+b^2+c^2}{2}=\frac{1}{2}\)
a) 3/4x16/9-7/5:(-21/20)
=4/3-(-4/3)
=8/3
b)=7/3-1/3x[-3/2+(2/3+2)]
=7/3-1/3x[-3/2+8/3]
=7/3-1/3x7/6
=7/3-7/18
=35/18
c)=(20+37/4):9/4
=117/4:9/4
=13
d)=6-14/5x25/8-8/5:1/4
=6-35/4-32/5
=-11/4-32/5
=-183/20
\(\left(x+5\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=-4\end{cases}}}\)
vậy x=-5 và x=-4
b) dễ tự làm
c)\(|x+9|-3=5\)
\(|x+9|=2\)
\(\Leftrightarrow\orbr{\begin{cases}x+9=2\\x+9=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-7\\x=7\end{cases}}}\)
vậy x=-7 hoặc x=7
1/3 công 2/5= 5/15 cộng với 6/15=11/15
NẾU ĐÚNG CHO MÌNH ĐÚNG NHÉ.
NẾU SAI CHO MÌNH SAI. CẢM ƠN CÁC BẠN. THANK
Áp dụng bđt bu nhi a, ta có \(M^2\le3\left(\frac{a}{b+c+2a}+...\right)\)
mà \(\frac{a}{b+c+2a}\le\frac{1}{4}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)\)
tương tự, ta có \(M^2\le\frac{3}{4}\left(\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{c}{c+b}\right)=\frac{9}{4}\)
=>\(M\le\frac{3}{2}\)
dấu = xảy ra <=> a=b=c
bn lên mạng hoặc vào câu hỏi tương tự nha!
chúc bn hok tốt!
hahaha!
#conmeo#
Vì a,b,c,d có vai trò như nhau
Giả sử \(a\ge b\ge c\ge d\)
=>\(a^2\ge b^2\ge c^2\ge d^2\)
=>\(\frac{1}{a^2}\le\frac{1}{b^2}\le\frac{1}{c^2}\le\frac{1}{d^2}\)
=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}\le\frac{1}{d^2}+\frac{1}{d^2}+\frac{1}{d^2}+\frac{1}{d^2}\)
=>\(1\le4.\frac{1}{d^2}\)
=>\(\frac{1}{4}\le\frac{1}{d^2}\)
=>\(4\ge d^2\)
=>\(2\ge d\)
Vì d là số tự nhiên khác 0
=>d=1,2
-Xét d=1
=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}=1\)
=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{1^2}=1\)
=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+1=1\)
=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=0\)
Vì\(\frac{1}{a^2}>0,\frac{1}{b^2}>0,\frac{1}{c^2}>0=>\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}>0\)
=>Vô lí
-Xét d=2
\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{2^2}=1\)
=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{4}=1\)
=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{3}{4}\)
Vì \(a\ge b\ge c\)
=>\(a^2\ge b^2\ge c^2\)
=>\(\frac{1}{a^2}\le\frac{1}{b^2}\le\frac{1}{c^2}\)
=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\le\frac{1}{c^2}+\frac{1}{c^2}+\frac{1}{c^2}\)
=>\(\frac{3}{4}\le3.\frac{1}{c^2}\)
=>\(\frac{1}{4}\le\frac{1}{c^2}\)
=>\(4\ge c^2\)
=>\(2\ge c\)
Vì \(c\ge d=>c\ge2\)
=>c=2
=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{3}{4}\)
=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{2^2}=\frac{3}{4}\)
=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{4}=\frac{3}{4}\)
=>\(\frac{1}{a^2}+\frac{1}{b^2}=\frac{2}{4}\)
Vì \(a\ge b\)
=>\(a^2\ge b^2\)
=>\(\frac{1}{a^2}\le\frac{1}{b^2}\)
=>\(\frac{1}{a^2}+\frac{1}{b^2}\le\frac{1}{b^2}+\frac{1}{b^2}\)
=>\(\frac{2}{4}\le\frac{2}{b^2}\)
=>\(\frac{1}{4}\le\frac{1}{b^2}\)
=>\(4\ge b^2\)
=>\(2\ge b\)
Vì \(b\ge c=>b\ge2\)
=>b=2
=>\(\frac{1}{a^2}+\frac{1}{b^2}=\frac{2}{4}\)
=>\(\frac{1}{a^2}+\frac{1}{2^2}=\frac{2}{4}\)
=>\(\frac{1}{a^2}+\frac{1}{4}=\frac{2}{4}\)
=>\(\frac{1}{a^2}=\frac{1}{4}\)
=>\(a^2=4=>a=2\)
Vậy a=2,b=2,c=2,d=2