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1) \(\Leftrightarrow\sqrt{\left(x+5\right)^2}=4\)
\(\Leftrightarrow\left|x+5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=4\\x+5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-9\end{matrix}\right.\)
2) \(ĐK:x\ge2\)
\(\Leftrightarrow\sqrt{x-2}=2\)
\(\Leftrightarrow x-2=4\Leftrightarrow x=6\left(tm\right)\)
3) \(\Leftrightarrow\left(x^2-x+4\right)-\sqrt{x^2-x+4}+\dfrac{1}{4}=\dfrac{9}{4}\)
\(\Leftrightarrow\left(\sqrt{x^2-x+4}-\dfrac{1}{2}\right)^2=\dfrac{9}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}-\dfrac{1}{2}=\dfrac{3}{2}\\\sqrt{x^2-x+4}-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}=2\\\sqrt{x^2-x+4}=-1\left(VLý\right)\end{matrix}\right.\)
\(\Leftrightarrow x^2-x+4=4\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
4) \(ĐK:x\ge0\)
\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{25}{4}\left(tm\right)\)
a:
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=3\)
=>|x-3|=3
=>x-3=3 hoặc x-3=-3
=>x=0 hoặc x=6
b: \(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}=2\)
=>\(\sqrt{\left(\sqrt{x-1}+1\right)^2}=2\)
=>\(\left|\sqrt{x-1}+1\right|=2\)
=>\(\left[{}\begin{matrix}\sqrt{x-1}+1=2\\\sqrt{x-1}+1=-2\left(loại\right)\end{matrix}\right.\Leftrightarrow\sqrt{x-1}=1\)
=>x-1=1
=>x=2
c:
ĐKXĐ: x>4/5
PT \(\Leftrightarrow\sqrt{\dfrac{5x-4}{x+2}}=2\)
=>\(\dfrac{5x-4}{x+2}=4\)
=>5x-4=4x+8
=>x=12(nhận)
d: ĐKXĐ: x-4>=0 và x+1>=0
=>x>=4
PT =>\(\left(\sqrt{x-4}+\sqrt{x+1}\right)^2=5^2=25\)
=>\(x-4+x+1+2\sqrt{\left(x-4\right)\left(x+1\right)}=25\)
=>\(\sqrt{4\left(x^2-3x-4\right)}=25-2x+3=28-2x\)
=>\(\sqrt{x^2-3x-4}=14-x\)
=>x<=14 và x^2-3x-4=(14-x)^2=x^2-28x+196
=>x<=14 và -3x-4=-28x+196
=>x<=14 và 25x=200
=>x=8(nhận)
a) \(\sqrt{x^2-6x+9}=3\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=3\)
\(\Leftrightarrow\left|x-3\right|=3 \)
TH1: \(\left|x-3\right|=x-3\) với \(x\ge3\)
Pt trở thành:
\(x-3=3\) (ĐK: \(x\ge3\))
\(\Leftrightarrow x=3+3\)
\(\Leftrightarrow x=6\left(tm\right)\)
TH2: \(\left|x-3\right|=-\left(x-3\right)\) với \(x< 3\)
Pt trở thành:
\(-\left(x-3\right)=3\) (ĐK: \(x< 3\))
\(\Leftrightarrow x-3=-3\)
\(\Leftrightarrow x=-3+3\)
\(\Leftrightarrow x=0\left(tm\right)\)
b) \(\sqrt{x+2\sqrt{x-1}}=2\) (ĐK: \(x\ge1\))
\(\Leftrightarrow x+2\sqrt{x-1}=4\)
\(\Leftrightarrow2\sqrt{x-1}=4-x\)
\(\Leftrightarrow4\left(x-1\right)=16-8x+x^2\)
\(\Leftrightarrow4x-4=16-8x+x^2\)
\(\Leftrightarrow x^2-12x+20=0\)
\(\Leftrightarrow\left(x-10\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=10\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
c) \(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\) (ĐK: \(x\ge\dfrac{4}{5}\))
\(\Leftrightarrow\dfrac{5x-4}{x+2}=4\)
\(\Leftrightarrow5x-4=4x+8\)
\(\Leftrightarrow x=12\left(tm\right)\)
1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)
\(\Leftrightarrow5-2x=36\)
\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)
2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)
\(\Leftrightarrow2-x=x+1\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)
3) \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x-2\)
\(\Leftrightarrow\left|x-5\right|=x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
Câu 1, \(\left(1\right)\hept{\begin{cases}\sqrt[4]{x^3}+\sqrt[5]{y^3}=35\\\sqrt[4]{x}+\sqrt[5]{y}=5\end{cases}}\)
ĐKXĐ: x > 0
Đặt \(\hept{\begin{cases}\sqrt[4]{x}=a\left(a\ge0\right)\\\sqrt[5]{y}=b\end{cases}}\)
Hệ ban đầu trở thành
\(\hept{\begin{cases}a^3+b^3=35\\a+b=5\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(a+b\right)\left(a^2-ab+b^2\right)=35\\a+b=5\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}5.\left[\left(a+b\right)^2-3ab\right]=35\\a+b=5\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(a+b\right)^2-3ab=7\\a+b=5\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}25-3ab=7\\a+b=5\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}ab=6\\a+b=5\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a\left(5-a\right)=6\\b=5-a\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}5a-a^2=6\\b=5-a\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a^2-5a+6=0\\b=5-a\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(a-3\right)\left(a-2\right)=0\\b=5-a\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a=3\\b=2\end{cases}\left(h\right)\hept{\begin{cases}a=2\\b=3\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt[4]{x}=3\\\sqrt[5]{y}=2\end{cases}}\left(h\right)\hept{\begin{cases}\sqrt[4]{x}=2\\\sqrt[5]{y}=3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=81\\y=32\end{cases}\left(h\right)\hept{\begin{cases}x=16\\y=243\end{cases}}}\)(Thỏa mãn)
Vậy
2/ Đặt \(\hept{\begin{cases}\sqrt{x}=a\ge0\\\sqrt{1-x}=b\ge0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a^3+b^3=a+2b\\a^2+b^2=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(a+b\right)\left(a^2+b^2-ab\right)=a+2b\\a^2+b^2=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(a+b\right)\left(1-ab\right)=a+2b\\a^2+b^2=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}b\left(a^2+ab+1\right)=0\\a^2+b^2=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}b=0\\a^2+b^2=1\end{cases}}\)
Bí
.\(đk:x\ge4\) \(x+\sqrt{x}+1+\left(2\sqrt{5}-1\right)\sqrt{x}=3x-2\sqrt{x-4}.\)
\(\Leftrightarrow(x-2\sqrt{5}.\sqrt{x}+5)+[(x-4)-2\sqrt{x-4}+1]=-3.\)
\(\Leftrightarrow[\sqrt{x}-\sqrt{5}]^2+[\sqrt{x-4}-1]^2=-3.\)
Phương trình vô nghiệm
1) \(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{4-2\sqrt{3}}=\sqrt{3}+1-\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}+1-\sqrt{3}+1=2\)
2) \(\dfrac{3}{5}\sqrt{25x-50}-\sqrt{x-2}=6\left(đk:x\ge2\right)\)
\(\Leftrightarrow3\sqrt{x-2}-\sqrt{x-2}=6\)
\(\Leftrightarrow2\sqrt{x-2}=6\)
\(\Leftrightarrow\sqrt{x-2}=3\)
\(\Leftrightarrow x-2=9\Leftrightarrow x=11\left(tm\right)\)
\(A=\dfrac{1-\sqrt{x}}{\sqrt{x}+2}=\dfrac{3-\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=\dfrac{3}{\sqrt{x}+2}-1\)
Có \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\)\(\Leftrightarrow\dfrac{3}{\sqrt{x}+2}-1\le\dfrac{1}{2}\)\(\Leftrightarrow A\le\dfrac{1}{2}\)
Dấu "=" xảy ra khi x=0 (tm)
Vậy \(A_{max}=\dfrac{1}{2}\)
Bài 2:
Đk: \(x\ge3;y\ge5;z\ge4\)
Pt\(\Leftrightarrow\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}+\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}+\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}=20\)
Áp dụng AM-GM có:
\(\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}\ge2\sqrt{\sqrt{x-3}.\dfrac{4}{\sqrt{x-3}}}=4\)
\(\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}\ge6\)
\(\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}\ge10\)
Cộng vế với vế \(\Rightarrow VT\ge20\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x-3}=\dfrac{4}{\sqrt{x-3}}\\\sqrt{y-5}=\dfrac{9}{\sqrt{y-5}}\\\sqrt{z-4}=\dfrac{25}{\sqrt{z-4}}\end{matrix}\right.\)\(\Leftrightarrow x=7;y=14;z=29\) (tm)
Vậy...
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\(\sqrt{5-x^2}+\sqrt{x^2+3}\le\sqrt{\left(1+1\right)\left(5-x^2+x^2+3\right)}=4\)
Dấu "=" xảy ra khi và chỉ khi \(5-x^2=x^2+3\)
\(\Leftrightarrow x^2=1\Rightarrow x=\pm1\)
Vậy nghiệm của pt là \(x=\pm1\)
cảm ơn anh > <