1,x=-10

2,x=10

Đáp án: x=-1 Giải thích các bước giải:

x(x-5)(x+5)-(x+2)( x 2 -2x+4 )=17

⇔x( x 2 +5x-5x-25)-( x 3 -2 x 2 +4x+2 x 2 -4x+8)=17

⇔ x 3 -25x - x 3 -8-17=0

⇔ x 3 -25x - x 3 -25=0

⇔-25x-25=0

⇔-25x=25

⇒x=-1

\(\Leftrightarrow x\left(x^2-25\right)-x^3-8=17\\ \Leftrightarrow x^3-25x-x^3=25\\ \Leftrightarrow-25x=25\Leftrightarrow x=-1\)

\(\Leftrightarrow x^3-25x-x^3-8=17\)

hay x=-1

Ta có: \(\left|x-4\right|-x+17=0\)

\(\Leftrightarrow\left|x-4\right|=x-17\)

\(\Leftrightarrow x-4=17-x\left(x< 4\right)\)

\(\Leftrightarrow2x=21\)

hay \(x=\dfrac{21}{2}\left(loại\right)\)

Đáp án cần chọn là: B

\(P=\dfrac{x}{\sqrt{x+y-x}}+\dfrac{y}{\sqrt{x+y-y}}=\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}\)

\(=\dfrac{x^2}{x\sqrt{y}}+\dfrac{y^2}{y\sqrt{x}}\ge\dfrac{\left(x+y\right)^2}{x\sqrt{y}+y\sqrt{x}}=\dfrac{\left(x+y\right)^2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}\)

\(\ge\dfrac{\left(x+y\right)^2}{\dfrac{x+y}{2}.\left(1.\sqrt{x}+1.\sqrt{y}\right)}\ge\dfrac{\left(x+y\right)^2}{\dfrac{x+y}{2}.\sqrt{\left(1^2+1^2\right)\left(x+y\right)}}=\dfrac{1}{\dfrac{1}{2}\sqrt{2}}=\sqrt{2}\)

"=" khi x = y = 1/2

giúp mình voi ah

Đặt \(t=\sqrt{x-3}\)\(\left(t\ge0\right)\)

\(\sqrt{8+t}+\sqrt{5-t}=5\)

\(\Leftrightarrow\left(\sqrt{8+t}+\sqrt{5-t}\right)^2=25\)

\(\Leftrightarrow8+t+5-t+2\sqrt{\left(8+t\right)\left(5-t\right)}=25\)

\(\Leftrightarrow2\sqrt{\left(8+t\right)\left(5-t\right)}=12\)

\(\Leftrightarrow\sqrt{\left(8+t\right)\left(5-t\right)}=6\)

\(\Leftrightarrow\left(8+t\right)\left(5-t\right)=36\)

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

\(\Leftrightarrow\orbr{\begin{cases}t=1\\t=-4\left(l\right)\end{cases}}\)

thay t=1 = căn (x-3) => x=4

điều kiện x-3 \(\ge0;5-\sqrt{x-3}\ge0\)(1)

đặt \(\sqrt{8+\sqrt{x-3}}=a\left(a\ge\sqrt{8}\right);\sqrt{5-\sqrt{x-3}}=b\left(b\ge0\right)\)

\(\hept{\begin{cases}a+b=5\\a^2+b^2=13\end{cases}< =>\hept{\begin{cases}a=5-b\\\left(5-b\right)^2+b^2=13\end{cases}< =>}}\)\(\hept{\begin{cases}a=5-b\\2b^2-10b+12=0\end{cases}< =>\hept{\begin{cases}a=3\\b=2\end{cases};\hept{\begin{cases}a=2\\b=3\end{cases}}}}\)

chỉ có a=3 là thoảm= mãn a \(\ge\sqrt{8}\)

\(\hept{\begin{cases}a=3\\b=2\end{cases}< =>\hept{\begin{cases}8+\sqrt{x-3}=9\\5-\sqrt{x-3}=4\end{cases}< =>x=4}}\)(thỏa mãn (1))

vậy x=4

\(\Leftrightarrow0\le\sqrt{3-2x}\le\sqrt{5}\\ \Leftrightarrow0\le3-2x\le5\\ \Leftrightarrow-1\le x\le\dfrac{3}{2}\)

giúp mình với . mình đang cần gấp nhé!