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\(A=\left(1+\frac{x^2}{y^2}\right)\left(1+\frac{y^2}{x^2}\right)\ge2\sqrt{\frac{x^2}{y^2}}.2\sqrt{\frac{y^2}{x^2}}=2.\frac{x}{y}.2.\frac{y}{x}=4\) ( Cosi )
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=1\)
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Đề thiếu x nguyên nhé bạn :)
\(x^2+10x+10=\left(x^2+10x+25\right)-15\)
Đặt \(x^2+10x+10=a^2\left(a\in Z\right)\)
Khi đó:\(\left(x+5\right)^2-a^2=15\)
\(\Leftrightarrow\left(x+5-a\right)\left(x+5+a\right)=15\)
Đến đây bạn lập ước ra ngay nhé ! Có điều hơi mệt tí,hihi !
sai rồi bạn. phải là \(a^2-\left(x+5\right)^2\)chứ
Bài 1:
\(\left(x-y+z\right)^2+\left(z-y\right)^2+\left(x-y+z\right)\left(2y-2z\right)\)
\(=\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)
\(=\left(x-y+z+y-z\right)^2\)
\(=x^2\)
Bài 2:
đk: \(x\ne\left\{0;-1;-2;-3;-4;-5\right\}\)
Xét BT trái ta có:
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+4\right)\left(x+5\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+4}-\frac{1}{x+5}\)
\(=\frac{1}{x}-\frac{1}{x+5}\)
\(=\frac{5}{x\left(x+5\right)}=\frac{5}{x^2+5x}\)
GT của biểu thức lớn sẽ là: \(\frac{5}{x^2+5x}\cdot\frac{x^2+5x}{5}=1\) không phụ thuộc vào biến
=> đpcm
Bài 1.
( x - y + z ) + ( z - y )2 + ( x - y + z )( 2y - 2z )
= ( x - y + z ) - 2( x - y + z )( z - y ) + ( z - y )2
= [ ( x - y + z ) - ( z - y ) ]2
= ( x - y + z - z + y )2
= x2
Bài 2. ĐKXĐ tự ghi nhé :))
\(\left(\frac{1}{x^2+x}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}\right)\times\left(\frac{x^2+5x}{5}\right)\)
\(=\left(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}\right)\times\left(\frac{x\left(x+5\right)}{5}\right)\)
\(=\left(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+4}-\frac{1}{x+5}\right)\times\left(\frac{x\left(x+5\right)}{5}\right)\)
\(=\left(\frac{1}{x}-\frac{1}{x+5}\right)\times\frac{x\left(x+5\right)}{5}\)
\(=\left(\frac{x+5}{x\left(x+5\right)}-\frac{x}{\left(x+5\right)}\right)\times\frac{x\left(x+5\right)}{5}\)
\(=\frac{x+5-x}{x\left(x+5\right)}\times\frac{x\left(x+5\right)}{5}\)
\(=\frac{5}{x\left(x+5\right)}\times\frac{x\left(x+5\right)}{5}=1\)
=> đpcm
3/
a/ \(A=\left(x-y\right)^2+\left(x+y\right)^2.\)
\(A=\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)\)
\(A=x^2-2xy+y^2+x^2+2xy+y^2\)
\(A=2x^2+2y^2\)
b/ \(B=\left(2a+b\right)^2-\left(2a-b\right)^2\)
\(B=\left(4a^2+4ab+b^2\right)-\left(4a^2-4ab+b^2\right)\)
\(B=4a^2+4ab+b^2-4a^2+4ab-b^2\)
\(B=8ab\)
c/ \(C=\left(x+y\right)^2-\left(x-y\right)^2\)
\(C=\left(x^2+2xy+y^2\right)-\left(x^2-2xy+y^2\right)\)
\(C=x^2+2xy+y^2-x^2+2xy-y^2\)
\(C=4xy\)
d/ \(D=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(D=\left(4x^2-4x+1\right)-2\left(4x^2-12x+9\right)+4\)
\(D=4x^2-4x+1-8x^2+24x-18+4\)
\(D=-4x^2+20x-13\)
\(A=\left(a+2b-5+b\right)^2-2ab+34=\left(a+2b-5\right)^2+2b\left(a+2b-5\right)+b^2-2ab+34\)
\(A=\left(a+2b-5\right)^2+5b^2-10b+5+29\)
\(A=\left(a+2b-5\right)^2+5\left(b-1\right)^2+29\ge29\)
\(A_{min}=29\) khi \(\hept{\begin{cases}a=3\\b=1\end{cases}}\)
\(B=x+\frac{25}{x}-8\ge2\sqrt{x.\frac{25}{x}}-8=2\)
\(B_{min}=2\) khi \(x=5\)
\(C=\frac{x^2-15x+36}{x}=x+\frac{36}{x}-15\ge2\sqrt{x.\frac{36}{x}}-15=-3\)
\(C_{min}=-3\) khi \(x=6\)
\(\left(2x+\frac{1}{x}\right)^2+\left(2y+\frac{1}{y}\right)^2\)
\(=\frac{\left(2x+\frac{1}{x}\right)^2}{1}+\frac{\left(2y+\frac{1}{y}\right)^2}{1}\)
\(\ge\frac{\left(2x+2y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)
\(\ge\frac{\left(2x+2y+\frac{4}{x+y}\right)^2}{2}=18\)
Đẳng thức xảy ra tại x=y=1/2
\(\frac{x^2-yz}{\left(x+y\right)\left(x+z\right)}+\frac{y^2-zx}{\left(y+z\right)\left(y+x\right)}+\frac{z^2-xy}{\left(z+x\right)\left(z+y\right)}=\frac{\left(x^2-yz\right)\left(y+z\right)+\left(y^2-zx\right)\left(x+z\right)+\left(z^2-xy\right)\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)
=\(\frac{x^2y+x^2z+xy^2+y^2z+xz^2+yz^2-x^2y-x^2z-xy^2-y^2z-xz^2-yz^2}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}=\frac{0}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}=0\)
lik.e nhé!
Đề này sửa dấu + thành trừ ở cái chỗ \(x^2-y^2\) nhé ! Còn thiếu dữ kiện của x,y là : \(x>1,y>1\) :
Ta có : \(P=\frac{\left(x^3+y^3\right)-\left(x^2+y^2\right)}{\left(x-1\right)\left(y-1\right)}\)
\(=\frac{\left(x^3-x^2\right)+\left(y^3-y^2\right)}{\left(x-1\right)\left(y-1\right)}\)
\(=\frac{x^2\left(x-1\right)}{\left(x-1\right)\left(y-1\right)}+\frac{y^2\left(y-1\right)}{\left(x-1\right)\left(y-1\right)}\)
\(=\frac{x^2}{y-1}+\frac{y^2}{x-1}\) \(\ge\frac{2xy}{\sqrt{\left(y-1\right)\left(x-1\right)}}\) ( Cô - si )
Lại có : \(\sqrt{\left(y-1\right)}=\sqrt{1\cdot\left(y-1\right)}\le\frac{1+y-1}{2}=\frac{y}{2}\)
Tương tự : \(\sqrt{x-1}\le\frac{x}{2}\)
\(\Rightarrow\sqrt{\left(y-1\right)\left(x-1\right)}\le\frac{xy}{4}\)
Khi đó : \(\frac{2xy}{\sqrt{\left(y-1\right)\left(x-1\right)}}\ge\frac{2xy}{\frac{xy}{4}}=8\)
hay : \(P\ge8\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{x^2}{y-1}=\frac{y^2}{x-1}\\y-1=1,x-1=1\end{cases}}\) \(\Leftrightarrow x=y=2\)
Vậy : \(min\) \(P=8\) tại \(x=y=2\)
sorry bạn