tính giá trị của biểu thức
P=(x+3)^2+(x-3).(x-3)-2.(x+2).(x-4) với x= -1/2
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\(a,P=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\left(x\ge0;x\ne1\right)\\ P=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(x+16\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\\ P=\dfrac{x+16}{\sqrt{x}+3}\\ b,P=4\Leftrightarrow\dfrac{x+16}{\sqrt{x}+3}=4\\ \Leftrightarrow x+16=4\sqrt{x}+12\\ \Leftrightarrow x-4\sqrt{x}+4=0\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\\ \Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
\(c,P=\dfrac{x+16}{\sqrt{x}+3}=\dfrac{x-9+25}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}\\ P=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\ge2\sqrt{\left(\sqrt{x}+3\right)\cdot\dfrac{25}{\sqrt{x}+3}}-6=2\cdot5-6=4\\ P_{min}=4\Leftrightarrow\left(\sqrt{x}+3\right)^2=25\Leftrightarrow\sqrt{x}+3=5\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow x=4\left(tm\right)\)
\(d,x=3-2\sqrt{2}\Leftrightarrow\sqrt{x}=\sqrt{2}-1\\ \Leftrightarrow P=\dfrac{3-2\sqrt{2}+16}{\sqrt{2}-1+3}=\dfrac{19-2\sqrt{2}}{\sqrt{2}+2}\\ P=\dfrac{\left(19-2\sqrt{2}\right)\left(2-\sqrt{2}\right)}{2}=\dfrac{42-23\sqrt{2}}{2}\)
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P=x^3+3/5x^2y-3xy-3/5x^2y-xy+x^3
=2x^3-4xy
=2*(-2)^3-4*(-2)*1/3
=-16+8/3=-40/3
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a)
Ta có: $2x^2+2y^2=5xy \Leftrightarrow 2\frac{x}{y}+\frac{y}{x}=5$
Đặt $t=\frac{x}{y}$, ta có $2t+\frac{1}{t}=5 \Rightarrow 2t^2-5t+1=0$
Giải phương trình trên ta được $t_1=\frac{1}{2}$ và $t_2=1$. Vì $0<x<y$ nên $t>0$, do đó $t=\frac{x}{y}=\frac{1}{2}$.
Từ đó suy ra $x=\frac{y}{2}$ và thay vào biểu thức $E$ ta được:
$E=\frac{x^2+y^2}{x^2-y^2}=\frac{\frac{y^2}{4}+y^2}{\frac{y^2}{4}-y^2}=-\frac{5}{3}$
Vậy kết quả là $E=-\frac{5}{3}$.
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Lời giải:
ĐKXĐ: $x>0$
a. \(P=\frac{x-1}{\sqrt{x}}:\left[\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+1)}+\frac{1-\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}\right]\)
\(=\frac{x-1}{\sqrt{x}}:\frac{x-1+1-\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}=\frac{x-1}{\sqrt{x}}:\frac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}+1)}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\frac{(\sqrt{x}+1)^2}{\sqrt{x}}\)
b.
\(x=\frac{4}{4+2\sqrt{3}}=(\frac{2}{\sqrt{3}+1})^2\Rightarrow \sqrt{x}=\frac{2}{\sqrt{3}+1}\)
\(P=\frac{(\frac{2}{\sqrt{3}+1}+1)^2}{\frac{2}{\sqrt{3}+1}}=\frac{3+3\sqrt{3}}{2}\)
a: Ta có: \(P=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\dfrac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x\ge0;x\ne4\)
\(P=\dfrac{x+\sqrt{x}}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{x-6\sqrt{x}+4}{x-4}\)
\(=\dfrac{\left(x+\sqrt{x}\right)\left(\sqrt{x}+2\right)-\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+x-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}+2x+x+2\sqrt{x}-\left(2x-4\sqrt{x}-\sqrt{x}+2\right)+x-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}+2x+x+2\sqrt{x}-2x+4\sqrt{x}+\sqrt{x}-2+x-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}+2x+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}\left(x+1\right)+2\left(x+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(x+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x+1}{\sqrt{x}-2}\)
Khi \(x=9+4\sqrt{5}\)
Ta có: \(4+4\sqrt{5}+5=2^2+2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2=\left(2+\sqrt{5}\right)^2\)
\(\Rightarrow\sqrt{x}=2+\sqrt{5}\)
\(\Rightarrow P=\dfrac{\left(2+\sqrt{5}\right)^2+1}{2+\sqrt{5}-2}=\dfrac{9+4\sqrt{5}+1}{\sqrt{5}}=\dfrac{10+4\sqrt{5}}{\sqrt{5}}=4+2\sqrt{5}\)
Vậy \(P=4+2\sqrt{5}\) khi \(x=9+4\sqrt{5}\).
\(D=\dfrac{x\sqrt{x}+2x+x+2\sqrt{x}-2x+4\sqrt{x}+\sqrt{x}-2+x-6\sqrt{x}+4}{x-4}\)
\(=\dfrac{x\sqrt{x}+2x+2}{x-4}\)
Khi x=9+4căn 5 thì \(D=\dfrac{\left(9+4\sqrt{5}\right)\left(\sqrt{5}+2\right)+2\sqrt{5}+4+2}{\sqrt{5}-2}\)
\(=\dfrac{9\sqrt{5}+18+20+8\sqrt{5}+2\sqrt{5}+6}{\sqrt{5}-2}\)
=(44+19căn 5)*(căn 5+2)
=44căn 5+88+95+38căn 5
=82căn 5+183
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(P=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)ĐK : \(x\ge0;x\ne4\)
Thay x = 9 vào P ta được
\(P=\dfrac{\sqrt{9}+3}{\sqrt{9}-2}=\dfrac{6}{1}=6\)
Với \(x>0;x\ne4\)
\(Q=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{x-4}\)
\(=\dfrac{x-3\sqrt{x}+2+5\sqrt{x}-2}{x-4}=\dfrac{x+2\sqrt{x}}{x-4}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
P=x2+6x+9+x2-6x+9-2x2+4x+16=4x+34
Thay x=-1/2 vào P ta có: P=4.(-1/2)+34=-2+34=32.kl.