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pascal bạn oi

Theo mk nghĩ thôi nhé, mk viết đáp số thôi nha

\(a,b,c=0\)

Trong 3 số a,b,c luôn tồn tại hai số cùng \(\ge\frac{1}{2}\) hoặc \(\le\frac{1}{2}\)Giả sử hai số đó là a và b

Ta có:\(c\left(2a-1\right)\left(2b-1\right)\ge0\Leftrightarrow c\left(4ab-2a-2b+1\right)\ge0\)

\(\Leftrightarrow4abc-2ac-2bc+c\ge0\Leftrightarrow4abc+c\ge2ac+2bc\)

Ta lại có:\(1=a^2+b^2+c^2+2abc\ge2ab+2abc+c^2\)

\(\Leftrightarrow1-c^2\ge2ab\left(c+1\right)\Leftrightarrow1-c\ge2ab\Leftrightarrow1\ge2ab+c\)\(\ge2\sqrt{2abc}\)

\(\Rightarrow1\ge8abc\Rightarrow abc\le\frac{1}{8}\).Từ \(a^2+b^2+c^2+2abc=1\Rightarrow\)

\(2+c=2a^2+2b^2+2c^2+4abc+c\)\(\ge2a^2+2b^2+2c^2+2ac+2bc\)

\(\Leftrightarrow1+1+c-a^2-b^2-c^2+2ab\ge a^2+b^2+c^2+2ab+2ac+2bc\)

\(\Leftrightarrow\left(a+b+c\right)^2\le1+2abc+c+2ab\le1+\frac{1}{4}+1=\frac{9}{4}\)

\(\Rightarrow a+b+c\le\frac{3}{2}\).Nên GTLN của M là \(\frac{3}{2}\) khi \(a=b=c=\frac{1}{2}\)

dễ

x² + y² + xy = x²y²

x² + xy + y² - x²y² = 0

4x² + 4xy + 4y² - 4x²y² = 0

( 4x² + 8xy + 4y² ) - ( 4x²y² + 8xy + 1 ) = -1       ( thêm - 1 )

( 2x + 2y )² - ( 2xy + 1 )² = -1

( 2x + 2y - 2xy - 1 ) ( 2x + 2y + 2xy + 1 ) = -1

\(\Rightarrow\)\(\hept{\begin{cases}2x+2y-2xy-1=1\\2x+2y+2xy+1=-1\end{cases}}\)hoặc \(\hept{\begin{cases}2x+2y-2xy-1=-1\\2x+2y+2xy+1=1\end{cases}}\)

suy ra tìm đc ( x; y ) \(\in\){ ( 0 ; 0 ) ; ( -1 ; 1 ) ; ( 1 ; -1 ) }

SKT-STT giúp mk bài tập này vs 

Tìm các số nguyên x dể bt \(A=\frac{x^5+1}{x^3+1}\)   có giá trị là số nguyên

máy tính mik khó viết nhưng bài này có mẫu chung nên dễ làm mà

bn cứ đưa mẫu ra có x-8 chung đó

sau đó tính tiếp theo bt là ra mà

bạn ơi bạn làm chi tiết ra ik mk thư rôi nhưng không đc

\(\frac{3}{a+2b}=\frac{1}{3}.\frac{9}{a+b+b}\le\frac{1}{3}.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\right)\)

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

\(\frac{3}{c+2a}\le\frac{1}{3}\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{a}\right)\)

Cộng theo vế ta được:

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

Số chính phương đó là 3136 . Hok tốt

câu a tự quy đồng cùng  mẫu rồi làm thôi :"))

b) \(\left[x.\left(x-1\right)\right].\left[\left(x-2\right).\left(x+1\right)\right]=24\)

\(\Leftrightarrow\left(x^2-x\right).\left(x^2-x-2\right)=24\)

Đặt \(x^2-x=k\), ta có:

\(k.\left(k-2\right)=24\)

\(\Leftrightarrow k^2-2k+1=25\)

\(\Leftrightarrow\left(k-1\right)^2=5^2\Leftrightarrow\orbr{\begin{cases}k-1=5\\k-1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}k=6\\k=-4\end{cases}}}\)

\(k=6\Rightarrow x^2-x=6\Rightarrow x^2-x-6=0\)

\(\Rightarrow x^2-3x+2x-6=0\Rightarrow x.\left(x-3\right)+2.\left(x-3\right)=0\)

\(\Rightarrow\left(x+2\right).\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)

\(k=-4\Rightarrow x^2-x+4=0\Rightarrow x^2-x+\frac{1}{4}+\frac{15}{4}=0\Rightarrow\left(x-\frac{1}{2}\right)^2=-\frac{15}{4}\left(\text{loại}\right)\)

c)\(x^4+2x^3+5x^2+4x-12=0\)

\(\Leftrightarrow x^4+2x^3+2x^2+4x+3x^2-12=0\)

\(\Leftrightarrow x^3.\left(x+2\right)+2x.\left(x+2\right)+3.\left(x^2-2^2\right)=0\)

\(\Leftrightarrow\left(x+2\right).\left(x^3+5x-6\right)=0\)

\(\Leftrightarrow\left(x+2\right).\left(x^3-x^2+x^2-x+6x-6\right)=0\)

\(\Leftrightarrow\left(x+2\right).\left[x^2.\left(x-1\right)+x.\left(x-1\right)+6.\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right).\left(x-1\right).\left(x^2+x+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}\text{vì }x^2+x+6>0\left(\text{tự c/m}\right)}\)

p/s: bn tự kết luận nha :))

a,\(B=\frac{x^5+x^2}{x^3-x^2+x}\left(ĐKXĐ:x\ne0\right)\)

\(\Rightarrow B=\frac{x^2\left(x^3+1\right)}{x\left(x^2-x+1\right)}=\frac{x\left(x+1\right)\left(x^2-x+1\right)}{x^2-x+1}=x^2+x\)

b,Để \(B=0\Rightarrow x^2+x=0\Leftrightarrow x\left(x+1\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

c,\(B=x^2+x=x^2+2.\frac{1}{2}x+\frac{1}{4}-\frac{1}{4}=\left(x+\frac{1}{2}\right)^2+\left(-\frac{1}{4}\right)\ge-\frac{1}{4}\)

Vậy MIn = -1/4 <=> x = -1/2