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Câu 1:
a,Bạn tự vẽ
b,Phương trình hoành độ giao điểm của (d1) và (d2) là:
\(\(\(-2x+3=x-1\Rightarrow-3x=-4\Rightarrow x=\frac{4}{3}\)\)\)
\(\(\(\Rightarrow y=\frac{4}{3}-1=\frac{1}{3}\)\)\)
Vậy tọa độ giao điểm của (d1) và (d2) là \(\(\(\left(\frac{4}{3};\frac{1}{3}\right)\)\)\)
c,Đường thẳng (d3) có dạng: y = ax + b
Vì (d3) song song với (d1) \(\(\(\Rightarrow\hept{\begin{cases}a=a'\\b\ne b'\end{cases}}\Rightarrow\hept{\begin{cases}a=-2\\b\ne3\end{cases}}\)\)\)
Khi đó (d3) có dạng: y = -2x + b
Vì (d3) đi qua điểm A( -2 ; 1) nên \(\(\(\Rightarrow x=-2;y=1\)\)\)
Thay x = -2 ; y = 1 vào (d3) ta được:\(\(\(1=-2.\left(-2\right)+b\Rightarrow b=-3\)\)\)
Vậy (d3) có phương trình: y = -2x - 3
Câu 2:
\(A=\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}\left(a>0;b>0;a\ne b\right)\)(Đề chắc phải như này)
\(\(\(=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}.\frac{\sqrt{a}-\sqrt{b}}{1}\)\)\)
\(\(\(=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\)\)\)
\(\(\(=\sqrt{a}^2-\sqrt{b}^2\)\)\)
\(\(\(=a-b\)\)\)
1/ ĐKXĐ: \(x\ge0,x\ne1\)
\(E=\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}-\dfrac{1}{\sqrt{x}-1}\right)-\left(1-\dfrac{\sqrt{x}}{x+1}\right)\)
= \(\left[\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}-1}\right]-\left(1+\dfrac{\sqrt{x}}{x+1}\right)\)
= \(\dfrac{2\sqrt{x}-x-1}{\left(x+1\right)\left(\sqrt{x}-1\right)}-\dfrac{x+1+\sqrt{x}}{x+1}\)
= \(\dfrac{-\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}-1\right)}-\dfrac{x+1+\sqrt{x}}{x+1}\)
= \(\dfrac{1-\sqrt{x}}{x+1}-\dfrac{x+1+\sqrt{x}}{x+1}\)
= \(\dfrac{1-\sqrt{x}-x-1-\sqrt{x}}{x+1}=\dfrac{-x-2\sqrt{x}}{x+1}\)
b/ Với \(x\ge0,x\ne1\)
Để \(E=-\dfrac{1}{7}\Leftrightarrow\dfrac{-x-2\sqrt{x}}{x+1}=-\dfrac{1}{7}\)
\(\Leftrightarrow-7x-14\sqrt{x}+x+1=0\)
\(\Leftrightarrow-6x-14\sqrt{x}+1=0\)
\(\Leftrightarrow\left(6\sqrt{x}+7-\sqrt{55}\right)\left(6\sqrt{x}+7+\sqrt{55}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}6\sqrt{x}+7-\sqrt{55}=0\\6\sqrt{x}+7+\sqrt{55}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{-7+\sqrt{55}}{6}\\\sqrt{x}=\dfrac{-7-\sqrt{55}}{6}\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{52-7\sqrt{55}}{18}\)
Vậy để \(E=-\dfrac{1}{7}\) thì \(x=\dfrac{52-7\sqrt{55}}{18}\)
Câu 2:
\(a,Đkxđ:x\ge0;x\ne9;x\ne4\)
\(b,P=\frac{2\sqrt{x}-9+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)-\left(x-9\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{2\sqrt{x}-9+2x-4\sqrt{x}+\sqrt{x}-2-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
Ta co:\(\Sigma\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}=\Sigma\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}\ge\frac{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)Ta lai co:
\(\Sigma x+\Sigma\frac{1}{x}=\Sigma\left(x+\frac{1}{4x}\right)+\frac{3}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3+\frac{3}{4}.\frac{9}{x+y+z}\ge3+\frac{3}{4}.\frac{9}{\frac{3}{2}}=\frac{15}{2}\)
Dau '=' xay ra khi \(x=y=z=\frac{1}{2}\)
Vay \(P_{min}=\frac{15}{2}\)khi \(x=y=z=\frac{1}{2}\)
1.
a. \(Q=\left(\dfrac{2\sqrt{a}}{\sqrt{a}+3}+\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3a+3}{a-9}\right):\left(\dfrac{2\sqrt{a}-2}{\sqrt{a}-3}-1\right)=\left(\dfrac{2\sqrt{a}}{\sqrt{a}+3}+\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3a+3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\right):\left(\dfrac{2\sqrt{a}-2-\sqrt{a}+3}{\sqrt{a}-3}\right)=\)
\(\dfrac{2\sqrt{a}\left(\sqrt{a}-3\right)+\sqrt{a}\left(\sqrt{a}+3\right)-3a-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}:\dfrac{\sqrt{a}+1}{\sqrt{a}-3}=\dfrac{2a-6\sqrt{a}+a+3\sqrt{a}-3a-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}.\dfrac{\sqrt{a}-3}{\sqrt{a}+1}=\dfrac{-3\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+1\right)}.\dfrac{\sqrt{a}-3}{\sqrt{a}+1}=\dfrac{-3\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}.\dfrac{\sqrt{a}-3}{\sqrt{a}+1}=-\dfrac{3}{\sqrt{a}+3}\)
Lewther
\(\dfrac{-\sqrt{x}-2}{\sqrt{x}-5}>-2\) (ĐKXĐ: \(x\ge0\) và \(x\ne25\))
\(\Leftrightarrow\dfrac{-\sqrt{x}-2}{\sqrt{x}-5}+2>0\Leftrightarrow\dfrac{-\sqrt{x}-2+2\left(\sqrt{x}-5\right)}{\sqrt{x}-5}>0\Rightarrow-\sqrt{x}-2+2\sqrt{x}-10>0\Leftrightarrow\sqrt{x}-12>0\Leftrightarrow\sqrt{x}>12\Leftrightarrow x>144\)
Vậy ...
Câu 1: ĐKXĐ: \(y\ge2\)
\(\Leftrightarrow\left\{{}\begin{matrix}6\left|2x-y\right|+3\sqrt{y-2}=15\\6\left|2x-y\right|-2\sqrt{y-2}=8\end{matrix}\right.\)
Trừ trên cho dưới ta được:
\(5\sqrt{y-2}=7\Leftrightarrow\sqrt{y-2}=\frac{7}{5}\Leftrightarrow y-2=\frac{49}{25}\Rightarrow y=\frac{99}{25}\)
Thay vào pt đầu:
\(2\left|2x-\frac{99}{25}\right|+\frac{7}{5}=5\Leftrightarrow\left|2x-\frac{99}{25}\right|=\frac{9}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{99}{5}=\frac{9}{5}\\2x-\frac{99}{5}=-\frac{9}{5}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{54}{5}\\x=9\end{matrix}\right.\)
Vậy hệ có 2 cặp nghiệm \(\left(x;y\right)=\left(\frac{54}{5};\frac{99}{5}\right);\left(9;\frac{99}{5}\right)\)
Câu 2:
Phương trình hoành độ giao điểm: \(x^2-\left(m-1\right)x-m^2-1=0\)
Ta có \(ac=-m^2-1< 0\) \(\forall m\Rightarrow\) pt luôn có 2 nghiệm trái dấu hay (d) luôn cắt (P) tại 2 điểm nằm về 2 phía trục tung
b/ Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=m-1\\x_1x_2=-m^2-1\end{matrix}\right.\)
\(\left|x_1\right|+\left|x_2\right|=2\sqrt{2}\Leftrightarrow x_1^2+x_2^2+2\left|x_1x_2\right|=8\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\left|x_1x_2\right|=8\)
\(\Leftrightarrow\left(m-1\right)^2-2\left(-m^2-1\right)+2\left|-m^2-1\right|=8\)
\(\Leftrightarrow5m^2-2m-3=0\Rightarrow\left[{}\begin{matrix}m=1\\m=-\frac{3}{5}\end{matrix}\right.\)