\(\Rightarrow x^2+y^2\ge2\sqrt{x^2y^2}=2xy\)

\(\Rightarrow1\ge2xy\)

\(\Rightarrow\frac{1}{2}\ge xy\)

Có \(x+y\ge2\sqrt{xy}\ge2\sqrt{\frac{1}{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)

Vậy \(Min_{x+y}=\sqrt{2}\)

Làm tương tự với max

Thêm đk: x,y>0

Tìm max:

Áp dụng BĐT bunhiacopxki ta có:

\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x+y\right)^2\)

\(\Leftrightarrow2\ge\left(x+y\right)^2\)

\(\Leftrightarrow\sqrt{2}\ge x+y\)

Dấu " = " xảy ra <=> x=y

KL:...............................

\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)ĐKXĐ : \(x\ne\pm4\)

\(\Leftrightarrow\frac{5\left(x^2-16\right)}{x^2-16}+\frac{96}{x^2-16}=\frac{\left(2x-1\right)\left(x-4\right)}{x^2-16}+\frac{\left(3x-1\right)\left(x+4\right)}{x^2-16}\)

\(\Leftrightarrow5x^2-80+96=2x^2-9x+4+3x^2+11x-4\)

\(\Leftrightarrow5x^2-2x^2-3x^2-11x+9x=4-4+80-96\)

\(\Leftrightarrow-2x=-16\)

\(\Leftrightarrow x=8\)( t/m )

Vậy....

\(\frac{x}{40}+\frac{x}{35}=7,5\)

\(\Rightarrow x\cdot35+x\cdot40=7,5\)

\(\Rightarrow x\cdot(35+40)=7,5\)

\(\Rightarrow x\cdot75=7,5\)

\(\Rightarrow x=0,1\)

P/S : Hoq chắc :>

Ta có : \(\frac{x}{40}+\frac{x}{35}=7,5\)

\(\Rightarrow\frac{35x}{1400}+\frac{40x}{1400}=\frac{15}{2}\)

\(\Rightarrow\frac{75x}{1400}=\frac{15}{2}\)

\(\Rightarrow\frac{3x}{56}=\frac{420}{56}\)

\(\Rightarrow3x=420\)

\(\Rightarrow x=420:3=140\)

Vậy x = 140

hướng dẫn cách làm-tự làm tiếp nha :)

a) đặt \(k=x^2-4x\), ta có:\(k^2-2k=15\)\(\Rightarrow k^2-2x+1=16\Rightarrow\left(k-1\right)^2=4^2=\left(-4\right)^2\)

b) đặt \(A=x^2-3x\), ta có: \(A^2-2A-8=0\Rightarrow A^2-2A+1=9\Rightarrow\left(A-1\right)^2=3^2=\left(-3\right)^2\)

c)theo đề \(\Leftrightarrow\orbr{\begin{cases}x^2-4x+3=0\\x^2-8x+9=0\end{cases}}\)

\(x^2-4x+3=0\Leftrightarrow x^2-4x+4=1\Leftrightarrow\left(x-2\right)^2=1^2=\left(-1\right)^2\)

\(x^2-8x+9=0\Leftrightarrow x^2-8x+16=7\Leftrightarrow\left(x-4\right)^2=\pm\sqrt{7}^2\)

vt ko chi tiết bn ib là đc rùi, sai tớ làm gì T.T 

mà tớ làm mẫu 1 bài thui nha, bài còn lại có cách làm òi. bn tự dựa vô nha

\(\text{Đặt }k=x^2-4x,\text{ta có:}\)

\(\left(x^2-4x\right)^2-2.\left(x^2-4x\right)=15\)

\(\Leftrightarrow k^2-2k=0\)

\(\Leftrightarrow k^2-2k+1=16\)

\(\Leftrightarrow\left(k-1\right)^2=16\)

\(\Leftrightarrow\orbr{\begin{cases}k-1=4\\k-1=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}k=5\\k=-3\end{cases}}}\)

\(\text{Với }k=5,\text{Ta có: }x^2-4x=5\Rightarrow x^2-4x-5=0\Rightarrow x^2-5x+x-5=0\)

\(\Rightarrow x.\left(x-5\right)+\left(x-5\right)=0\Rightarrow\left(x+1\right).\left(x-5\right)=0\Rightarrow\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)

\(\text{Với }k=-3,\text{ta có: }x^2-4x=-3\Rightarrow x^2-4x+3=0\Rightarrow k^2-3x-x+3=0\)

\(\Rightarrow x.\left(x-3\right)-\left(x-3\right)=0\Rightarrow\left(x-1\right).\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)

Vậy...