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a)\(đkx\ge1,x\ne-1\)
\(\sqrt{\dfrac{x-1}{x+1}}=2\)
\(\Leftrightarrow\dfrac{x-1}{x+1}=4\)
\(\Leftrightarrow x-1=4x-4\)
\(\Leftrightarrow x=1\)(nhận)
Vậy S=\(\left\{1\right\}\)
c)đk\(25x^2-10x+1=\) \(\left(5x-1\right)^2\ge0\Leftrightarrow x\ge\dfrac{1}{5}\)
\(\sqrt{25x^2-10x+1}+2x=1\)
\(\Leftrightarrow\sqrt{\left(5x-1\right)^2}+2x=1\)
\(\Leftrightarrow5x-1+2x=1\)
\(\Leftrightarrow x=\dfrac{2}{7}\)(nhận)
Vậy S=\(\left\{\dfrac{2}{7}\right\}\)
c: Ta có: \(\sqrt{25x^2-10x+1}+2x=1\)
\(\Leftrightarrow\left|5x-1\right|=1-2x\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=1-2x\left(x\ge\dfrac{1}{5}\right)\\5x-1=2x-1\left(x< \dfrac{1}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{7}\left(nhận\right)\\x=0\left(nhận\right)\end{matrix}\right.\)
1.
Ta có: $4x^2+4x+3=(4x^2+4x+1)+2=(2x+1)^2+2\geq 0+2=2$
$\Rightarrow A=\frac{6}{4x^2+4x+3}\leq \frac{6}{2}=3$
Vậy $A_{\max}=3$. Giá trị này đạt tại $2x+1=0\Leftrightarrow x=\frac{-1}{2}$
2.
$6+4x+x^2=(x^2+4x+4)+2=(x+2)^2+2\geq 0+2=2$
$\Rightarrow \frac{4}{6+4x+x^2}\leq \frac{4}{2}=2$
$\Rightarrow \frac{-4}{6+4x+x^2}\geq -2$
$\Rightarrow B\geq -2$
Vậy $B_{\min}=-2$. Giá trị này đạt tại $x+2=0\Leftrightarrow x=-2$
a: Xét (O) có
MA là tiếp tuyến
MB là tiếp tuyến
Do đó: MA=MB
hay M nằm trên đường trung trực của AB(1)
Ta có: OA=OB
nên O nằm trên đường trung trực của AB(2)
Từ (1) và (2) suy ra OM⊥AB
bài 7
A=\(\dfrac{x+2}{\sqrt{x^3}-1}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)}+\dfrac{-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
A=\(\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
A=\(\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)=\(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+x+1\right)}\)
A=\(\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
bài 8
P=\(\left[\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)^2}\right].\dfrac{\left(x-1\right)^2}{4x}\)
P=\(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}.\dfrac{\left(x-1\right)^2}{4x}\)
P=\(\dfrac{2\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\dfrac{\left(x-1\right)^2}{4x}\)=\(\dfrac{x-1}{2\sqrt{x}\left(\sqrt{x}-1\right)}\)
P=\(\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)
bài 9
P=\(\left[\dfrac{2\sqrt{xy}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}\right].\dfrac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
P=\(\dfrac{4\sqrt{xy}-\left(\sqrt{x}+\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}.\dfrac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
P=\(\dfrac{2\sqrt{xy}-x-y}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}.\dfrac{\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
P=\(\dfrac{-\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}.\dfrac{\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
P=\(\dfrac{-\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)
bài 10
P=\(\left[\dfrac{1}{\sqrt{x}+2}-\dfrac{2}{\left(\sqrt{x}+2\right)^2}\right]:\left[\dfrac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{1}{\sqrt{x}-2}\right]\)
P=\(\dfrac{\sqrt{x}+2-2}{\left(\sqrt{x}+2\right)^2}:\dfrac{2-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P=\(\dfrac{\sqrt{x}}{\left(\sqrt{x}+2\right)^2}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{-\sqrt{x}}\)=\(\dfrac{-\left(\sqrt{x}-2\right)}{\sqrt{x}+2}\)
thay \(x=3-2\sqrt{2}\) vào P ta có:
\(\dfrac{x+8}{\sqrt{x}+1}=\dfrac{3-2\sqrt{2}+8}{\sqrt{3-2\sqrt{2}}+1}=\dfrac{11-2\sqrt{2}}{\sqrt{2}-1+1}=\dfrac{11-2\sqrt{2}}{\sqrt{2}}\)
\(b,x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\)
Thay vào P, ta được:
\(P=\dfrac{3-2\sqrt{2}+8}{\sqrt{\left(\sqrt{2}-1\right)^2}+1}=\dfrac{11-2\sqrt{2}}{\sqrt{2}}=\dfrac{11\sqrt{2}-4}{2}\)
Bài 2:
e) \(\sqrt{4x-8}-12\sqrt{\dfrac{x-2}{9}}=\sqrt{x-2}-12\left(đk:x\ge2\right)\)
\(\Leftrightarrow\sqrt{4}.\sqrt{x-2}-12.\sqrt{\dfrac{1}{9}}.\sqrt{x-2}=\sqrt{x-2}-12\)
\(\Leftrightarrow2\sqrt{x-2}-4\sqrt{x-2}=\sqrt{x-2}-12\)
\(\Leftrightarrow3\sqrt{x-2}=12\)
\(\Leftrightarrow\sqrt{x-2}=4\)
\(\Leftrightarrow x-2=16\Leftrightarrow x=18\left(tm\right)\)
b) Gọi (d3): y=ax+b
Vì (d3)//(d1) nên \(a=-\dfrac{2}{3}\)
Vậy: (d3): \(y=\dfrac{-2}{3}x+b\)
Thay x=6 vào (d2), ta được:
\(y=-2\cdot6+4=-12+4=-8\)
Thay x=6 và y=-8 vào (d3), ta được:
\(\dfrac{-2}{3}\cdot6+b=-8\)
\(\Leftrightarrow b=-4\)
Vậy: (d3): \(y=\dfrac{-2}{3}x-4\)
