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Với 1 ≤ x < 2
A = (x + 3)/2
Với x ≥ 2
A = (x + 3)/[2√(x - 1)]
b/ Xét 1 ≤ x < 2
A ≥ (3 + 1)/2 = 2
Xét x ≥ 2
A = 2 + [√(x - 1) - 2]²/[2√(x - 2)] ≥ 2
Kết hợp 2 TH thì min là 2 khi x = 1 hoặc x = 5
\(P=\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{4-6\sqrt{a}}{1-a}-\frac{-3}{\sqrt{a}+1}\)
ĐK : \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)
a) \(P=\frac{\sqrt{a}}{\sqrt{a}-1}+\frac{4-6\sqrt{a}}{a-1}+\frac{3}{\sqrt{a}+1}\)
\(=\frac{\sqrt{a}}{\sqrt{a}-1}+\frac{4-6\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{3}{\sqrt{a}+1}\)
\(=\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{4-6\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{3\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\frac{a+\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{4-6\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{3\sqrt{a}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\frac{a+\sqrt{a}+4-6\sqrt{a}+3\sqrt{a}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\frac{a-2\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\frac{\sqrt{a}-1}{\sqrt{a}+1}\)
Với \(a=4-2\sqrt{3}\)( tmđk \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\))
\(P=\frac{\sqrt{4-2\sqrt{3}}-1}{\sqrt{4-2\sqrt{3}}+1}\)
\(=\frac{\sqrt{3-2\sqrt{3}+1}-1}{\sqrt{3-2\sqrt{3}+1}+1}\)
\(=\frac{\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}+1^2}-1}{\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}+1^2}+1}\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}-1}{\sqrt{\left(\sqrt{3}-1\right)^2}+1}\)
\(=\frac{\left|\sqrt{3}-1\right|-1}{\left|\sqrt{3}-1\right|+1}\)
\(=\frac{\sqrt{3}-1-1}{\sqrt{3}-1+1}=\frac{\sqrt{3}-2}{\sqrt{3}}\)
b) \(P=\frac{\sqrt{a}-1}{\sqrt{a}+1}=\frac{\sqrt{a}+1-2}{\sqrt{a}+1}=1-\frac{2}{\sqrt{a}+1}\)( ĐK \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\))
Để P đạt giá trị nguyên => \(\frac{2}{\sqrt{a}+1}\)nguyên
=> \(2⋮\sqrt{a}+1\)
=> \(\sqrt{a}+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
=> \(\sqrt{a}\in\left\{0;1\right\}\)< đã loại hai trường hợp âm >
=> \(a\in\left\{0\right\}\)< loại trường hợp a = 1 >
Vậy với a = 0 thì P có giá trị nguyên
a)
\(A=\frac{\sqrt{a}+3}{\sqrt{a}-2}-\frac{\sqrt{a}-1}{\sqrt{a}+2}+\frac{4\sqrt{a}-4}{4-a}\)
\(=\frac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)-\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}+\frac{4\sqrt{a}-4}{4-\sqrt{a}}\)
\(=\frac{a+2\sqrt{a}+3\sqrt{a}+6-a-2\sqrt{a}-\sqrt{a}+2}{a-4}+\frac{4\sqrt{a}-4}{4-a}\)
\(=\frac{a-a+\left(2+3-2-1\right)\sqrt{a}+6+2}{a-4}+\frac{-4\sqrt{a}+4}{a-4}\)
\(=\frac{2\sqrt{a}+8}{a-4}+\frac{-4\sqrt{a}+4}{a-4}\)
\(=\frac{2\sqrt{a}+8-4\sqrt{a}+4}{\left(a-4\right)^2}\)
\(=\frac{-2\sqrt{a}+12}{\left(a-4\right)^2}\)
b) thấy A = 9 vào biểu thức , ta có :
\(9=\frac{-2\sqrt{a}+12}{\left(a-4\right)^2}\)
\(< =>\frac{9\left(a-4\right)^2}{\left(a-4\right)^2}=\frac{-2\sqrt{a}+12}{\left(a-4\right)^2}\)
\(< =>9\left(a-4\right)^2=-2\sqrt{a}+12\)
\(< =>9.\left(a^2-2a.4+4^2\right)=-2\sqrt{a}+12\)
\(< =>9a^2-72a+144=-2\sqrt{a}+12\)
\(< =>9a^2-72a+2\sqrt{a}=12-144\)
\(< =>\sqrt{a}\left(9\sqrt{a}^3-72\sqrt{a}+2\right)=-132\)
\(\)
TỚI ĐÂY AI BIẾT THÌ GIẢI TIẾP NHA , MÌNH HẾT BIẾT CÁCH LÀM RỒI
a) \(\sqrt{x}+\sqrt{\frac{x}{9}}-\frac{1}{3}\sqrt{4x}=5\)
ĐK : x ≥ 0
<=>\(\sqrt{x}+\sqrt{x\times\frac{1}{9}}-\frac{1}{3}\sqrt{2^2x}=5\)
<=> \(\sqrt{x}+\sqrt{x\times\left(\frac{1}{3}\right)^2}-\left(\frac{1}{3}\times\left|2\right|\right)\sqrt{x}=5\)
<=> \(\sqrt{x}+\left|\frac{1}{3}\right|\sqrt{x}-\left(\frac{1}{3}\times2\right)\sqrt{x}=5\)
<=> \(\sqrt{x}+\frac{1}{3}\sqrt{x}-\frac{2}{3}\sqrt{x}=5\)
<=> \(\sqrt{x}\left(1+\frac{1}{3}-\frac{2}{3}\right)=5\)
<=> \(\sqrt{x}\times\frac{2}{3}=5\)
<=> \(\sqrt{x}=\frac{15}{2}\)
<=> \(x=\frac{225}{4}\)( tm )
a, \(\sqrt{x^2-4x+4}=3\Leftrightarrow\sqrt{\left(x-2\right)^2}=3\)
\(\Leftrightarrow x-2=3\Leftrightarrow x=5\)
b, \(\sqrt{x^2-10x+25}=x+3\Leftrightarrow\sqrt{\left(x-5\right)^2}=x+3\)
\(\Leftrightarrow x-5=x+3\Leftrightarrow0\ne8\)( vô nghiệm )
a, \(P=\frac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+1=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+1\)
\(=\sqrt{x}\left(\sqrt{x}+1\right)-2\sqrt{x}-1+1=x+\sqrt{x}-2\sqrt{x}=x-\sqrt{x}\)
b, \(P=x-\sqrt{x}=x-\sqrt{x}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge\frac{-1}{4}\)
Vậy Min P =-1/4
c, Chắc bằng nhau vì cùng dương mà
Phần a như bạn Đỗ Ngọc Hải chỉ thêm ĐKXĐ : x >= 0
b) Đkxd X >=0
Ta Có P = x-\(\sqrt{x}\) -2√x.½+1/4 -1/4=\(\left(\sqrt{x}-\frac{1}{2}\right)^2\)\(-\frac{1}{4}\)
Có √x>=0<=> (√x-½)2>=1/4<=>(√x-½)2-1/4>=0=>P>=0
Hay min p =0
Dấu = xảy ra <=> x=0
Vậy để minP=0<=>x=0
C)Dkxd x>1
CóP>=0(chứng minh trên )
=>|P|=P
\(A=\frac{\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(A=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{4}{x-1}\)
b) \(\frac{4}{x-1}=7\)
\(\Leftrightarrow4=7.\left(x-1\right)\)
\(\Leftrightarrow\frac{4}{7}=x-1\)
\(\Leftrightarrow\frac{4}{7}+1=x\)
\(\Leftrightarrow\frac{11}{7}=x\)
\(\Rightarrow x=\frac{11}{7}\)
