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mk giải 1 bài lm mẩu nha .
+) ta có : \(A=x-12\sqrt{x}\Leftrightarrow x-12\sqrt{x}-A=0\)
vì phương trình này luôn có nghiệm \(\Leftrightarrow\Delta'\ge0\)
\(\Leftrightarrow6^2+A\ge0\Leftrightarrow A\ge-36\)
vậy giá trị nhỏ nhất của \(A\) là \(-36\) dấu "=" xảy ra khi \(\sqrt{x}=\dfrac{-b'}{a}=\dfrac{6}{1}=6\Leftrightarrow x=36\)
mấy câu còn lại bn chuyển quế đưa về phương trình bật 2 theo \(x\) rồi giải như trên là đc :
lộn ! là phương trình bật 2 đối với ẩn là \(\sqrt{x}\) nha :
DƯƠNG PHAN KHÁNH DƯƠNG
ĐKXĐ : \(x\ge0\) và \(x\ne\dfrac{1}{9}\)
\(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}-1}=\dfrac{6}{5}\)
\(\Leftrightarrow\dfrac{5\sqrt{x}\left(\sqrt{x}+1\right)}{5\left(3\sqrt{x}-1\right)}=\dfrac{6\left(3\sqrt{x}-1\right)}{5\left(3\sqrt{x}-1\right)}\)
\(\Leftrightarrow5\sqrt{x}\left(\sqrt{x}+1\right)=6\left(3\sqrt{x}-1\right)\)
\(\Leftrightarrow5x+5\sqrt{x}-18\sqrt{x}+6=0\)
\(\Leftrightarrow5x-13\sqrt{x}+6=0\)
\(\Leftrightarrow5x-10\sqrt{x}-3\sqrt{x}+6=0\)
\(\Leftrightarrow5\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(5\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2=0\\5\sqrt{x}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{9}{25}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{9}{25};4\right\}\)
Học tốt !
Câu 1:
a: \(P=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b: Để P<1 thì \(\dfrac{\sqrt{a}-4-\sqrt{a}+2}{\sqrt{a}-2}< 0\)
\(\Leftrightarrow\sqrt{a}-2< 0\)
hay 0<a<4
a) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\)
\(=\dfrac{4a^2b^3}{8\sqrt{2}a^3b^3}\)
\(=\dfrac{1}{2\sqrt{2}a}\)
\(=\dfrac{\sqrt{2}}{4a}\)
b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
chịu đấy :v
c) \(\sqrt{\dfrac{\left(x-2\right)^2}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{3-x}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{-\left(x-3\right)}+\dfrac{x^2-1}{x-3}\)
\(=-\dfrac{x-2}{x-3}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{-\left(x-2\right)+x^2-1}{x-3}\)
\(=\dfrac{-x+1+x^2}{x-3}\)
d) \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1^2\right)}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(x-1\right)^2}\)
\(=\dfrac{1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{x-1}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(\sqrt{y}-1\right)\left(x-1\right)}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{x\sqrt{y}-\sqrt{y}-x+1}\)
e) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\dfrac{\sqrt{x^2\cdot\left(x+2\right)}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\sqrt{x^2}\)
\(=4x-2\sqrt{x}+x\)
\(=5x-2\sqrt{2}\)
a) \(\sqrt{\dfrac{x-2\sqrt{x+1}}{x+2\sqrt{x+1}}}\) = \(\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}\) = \(\dfrac{\sqrt{x-1}}{\sqrt{x+1}}\)
b) \(\dfrac{x-1}{\sqrt{y}-1}\)\(\sqrt{\dfrac{y-2\sqrt{y+1}}{\left(x-1\right)^4}}\)
= \(\dfrac{x-1}{\sqrt{y}-1}\) \(\sqrt{\dfrac{\left(y-1\right)^4}{\left(x-1\right)^4}}\)
= \(\dfrac{x-1}{\sqrt{y}-1}\)\(\dfrac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^2}\)
= \(\dfrac{\sqrt{y-1}}{x-1}\)
Chúc bạn học tốt :3
\(E=\frac{x+\sqrt{x}}{x-2\sqrt{x}+1}:\left(\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{1}{1-\sqrt{x}}+\frac{2-x}{x-\sqrt{x}}\right)\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\) \(\left[\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\sqrt{x}}+\frac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\)\(\left[\frac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(E=\frac{x}{\sqrt{x}-1}\)
b) \(E>1\Leftrightarrow\frac{x}{\sqrt{x}-1}>1\)
\(\Leftrightarrow\frac{x}{\sqrt{x}-1}-1>0\)
\(\Leftrightarrow\frac{x}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\frac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\frac{x-2\sqrt{x}+1+\sqrt{x}}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\frac{\left(\sqrt{x}-1\right)^2+\sqrt{x}}{\sqrt{x}-1}>0\)
\(\Rightarrow\sqrt{x}-1>0\) vì tử của phân số luôn \(\ge0\forall x\ge0\)
\(\Rightarrow x>1\)
kết hợp với ĐKXĐ \(x\ge0\Rightarrow x>1\)
vậy \(x>1\) thì \(E>1\)
\(a.P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)
Để : \(P\in Z\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\in Z\Leftrightarrow\left(\sqrt{x}+1\right)\in\left\{\pm1;\pm2\right\}\)
+) \(\sqrt{x}+1=1\Leftrightarrow x=0\left(TM\right)\)
+) \(\sqrt{x}+1=-1\Leftrightarrow vô-n^o\)
+) \(\sqrt{x}+1=2\Leftrightarrow x=1\left(KTM\right)\)
+) \(\sqrt{x}+1=-2\Leftrightarrow vô-n^o\)
KL.............
\(b.Q=\dfrac{\sqrt{a}+1}{\sqrt{a}+2}=\dfrac{\sqrt{a}+2-1}{\sqrt{a}+2}=1-\dfrac{1}{\sqrt{a}+2}\)
Để : \(Q\in Z\Leftrightarrow\dfrac{1}{\sqrt{a}+2}\in Z\Leftrightarrow\left(\sqrt{a}+2\right)\in\left\{\pm1\right\}\)
+) \(\sqrt{a}+2=1\Leftrightarrow vô-n^o\)
+) \(\sqrt{a}+2=-1\Leftrightarrow vô-n^o\)
KL............
\(c.A=\dfrac{\sqrt{a}-1}{\sqrt{a}-4}=\dfrac{\sqrt{a}-4+3}{\sqrt{a}-4}=1+\dfrac{3}{\sqrt{a}-4}\)
Để : \(A\in Z\Leftrightarrow\dfrac{3}{\sqrt{a}-4}\in Z\Leftrightarrow\left(\sqrt{a}-4\right)\in\left\{\pm1;\pm3\right\}\)
+) \(\sqrt{a}-4=1\Leftrightarrow a=25\left(TM\right)\)
+) \(\sqrt{a}-4=-1\Leftrightarrow a=9\left(TM\right)\)
+) \(\sqrt{a}-4=3\Leftrightarrow a=49\left(TM\right)\)
+) \(\sqrt{a}-4=-3\Leftrightarrow a=1\left(TM\right)\)
KL............
P/s : Mình thấy đề bài b sai nhé , mẫu phải là \(\sqrt{a}-2\) thì mới phù hợp ĐK đã cho .
Vì x\(\ge0,x\ne1\) nên \(\sqrt{x}+1\ge1\), do đó, để P>0 thì\(-x+6\sqrt{x}+9>0\Leftrightarrow-\left(x-6\sqrt{x}+9\right)+18>0\Leftrightarrow-\left(\sqrt{x}-3\right)^2>-18\Leftrightarrow\left(\sqrt{x}-3\right)^2< 18\Leftrightarrow\sqrt{\left(\sqrt{x}-3\right)^2}< \sqrt{18}\Leftrightarrow\left|\sqrt{x}-3\right|< 3\sqrt{2}\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3< 3\sqrt{2}\\-\left(\sqrt{x}-3\right)< 3\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}< 3\sqrt{2}+3\\\sqrt{x}>3-3\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< \left(3\sqrt{2}+3\right)^2\\x>\left(3-3\sqrt{2}\right)^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< 18+9+18\sqrt{2}\\x>9+18-18\sqrt{2}\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x< 27+18\sqrt{2}\\x>27-18\sqrt{2}\end{matrix}\right.\Leftrightarrow27-18\sqrt{2}< x< 27+18\sqrt{2}\)
Ơ, sao ko tải đc, đáp án cuối cùng là \(27-18\sqrt{2}< x< 27+18\sqrt{2}\)