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Chị tag ko dính em chút nào cả :( Em thứ nhá
b) ĐK: \(x\ge1\).Đặt \(\sqrt{x-1}=a\ge0;\sqrt{x+1}=b\ge0\). Nhưng do \(x\ge1\) nên \(b\ge\sqrt{2}\)
\(PT\Leftrightarrow3a+\frac{1}{3}b=2\sqrt{ab}\) \(\Leftrightarrow9a+b=6\sqrt{ab}\) (1)
Mặt khác ta có \(b^2-a^2=2\) (2). Từ (1) và (2) ta có:
\(\left\{{}\begin{matrix}9ab+b=6\sqrt{ab}\\\left(b-a\right)\left(a+b\right)=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}81a^2b^2+18ab^2+b^2=36ab\\b^2-a^2=2\end{matrix}\right.\)
Xét pt (1) của hệ: \(\Leftrightarrow b\left(81a^2b+18ab+b-36a\right)=0\)
b = 0 => loại
\(81a^2b+18ab+b-36a=0\Leftrightarrow b\left(81a^2+18a+1\right)=36a\)
Dễ thấy \(81a^2+18a+1>0\forall a\ge0\)
\(b=\frac{36a}{81a^2+18a+1}\). Thay vào pt thứ 2 của hệ suy ra \(\left(\frac{36a}{81a^2+18a+1}\right)^2-a^2-2=0\)
Và bí -_-"
a, ĐKXĐ: \(x\ge-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{3}{2}.2\sqrt{1+3x}-\dfrac{5}{3}.3\sqrt{1+3x}-\dfrac{1}{4}.4\sqrt{1+3x}=1\\ \Leftrightarrow3\sqrt{1+3x}-5\sqrt{1+3x}-\sqrt{1+3x}=1\\ \Leftrightarrow-3\sqrt{1+3x}=1\\ \Leftrightarrow\sqrt{1+3x}=-\dfrac{1}{3}\left(vô.lí\right)\)
b, \(\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=3\\ \Leftrightarrow\left|x-\dfrac{1}{2}\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=3\\x-\dfrac{1}{2}=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
a) ĐKXĐ: \(x\ge-\dfrac{1}{3}\)
\(pt\Leftrightarrow3\sqrt{3x+1}-5\sqrt{3x+1}-\sqrt{3x+1}=1\)
\(\Leftrightarrow-3\sqrt{3x+1}=1\Leftrightarrow\sqrt{3x+1}=-\dfrac{1}{3}\left(VLý\right)\)
Vậy \(S=\varnothing\)
b) \(pt\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=3\Leftrightarrow\left|x-\dfrac{1}{2}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=3\\x-\dfrac{1}{2}=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
\(PT\Leftrightarrow\left|3x-2\right|=8\Leftrightarrow\left[{}\begin{matrix}3x-2=8\\2-3x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=-2\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{\left(3x-2\right)^2}=8\)
\(\Leftrightarrow\left|3x-2\right|=8\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=8\\3x-2=-8\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=-2\end{matrix}\right.\)
2: ĐKXĐ: x>=0
\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)
=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)
=>\(\sqrt{3x}-4\sqrt{3x}+\sqrt{3x}=-4\)
=>\(-2\sqrt{3x}=-4\)
=>\(\sqrt{3x}=2\)
=>3x=4
=>\(x=\dfrac{4}{3}\left(nhận\right)\)
3:
ĐKXĐ: x>=0
\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)
=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)
=>\(13\sqrt{2x}=20+3\sqrt{2}\)
=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)
=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)
=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)
4: ĐKXĐ: x>=-1
\(\sqrt{16x+16}-\sqrt{9x+9}=1\)
=>\(4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>\(\sqrt{x+1}=1\)
=>x+1=1
=>x=0(nhận)
5: ĐKXĐ: x<=1/3
\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)
=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)
=>\(5\sqrt{1-3x}=10\)
=>\(\sqrt{1-3x}=2\)
=>1-3x=4
=>3x=1-4=-3
=>x=-3/3=-1(nhận)
6: ĐKXĐ: x>=3
\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\left(\dfrac{2}{3}+\dfrac{1}{6}-1\right)=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)
=>x-3=16
=>x=19(nhận)
a) ĐKXĐ: \(x\ge0\)
Ta có: \(3\sqrt{18x}-5\sqrt{8x}+4\sqrt{50x}=38\)
\(\Leftrightarrow9\sqrt{2x}-10\sqrt{2x}+20\sqrt{2x}=38\)
\(\Leftrightarrow19\sqrt{2x}=38\)
\(\Leftrightarrow\sqrt{2x}=2\)
\(\Leftrightarrow2x=4\)
hay x=2(thỏa ĐK)
b) ĐKXĐ: \(x\ge0\)
Ta có: \(3\sqrt{12x}-2\sqrt{27x}+4\sqrt{3x}=8\)
\(\Leftrightarrow6\sqrt{3x}-6\sqrt{3x}+4\sqrt{3x}=8\)
\(\Leftrightarrow\sqrt{3x}=2\)
\(\Leftrightarrow3x=4\)
hay \(x=\dfrac{4}{3}\)
c) ĐKXĐ: \(x\ge5\)
Ta có: \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
hay x=9
a)
\(3.3\sqrt{2x}-5.2\sqrt{2x}+4.5.\sqrt{2x}=38\\ \Leftrightarrow19\sqrt{2x}=38\\ \Leftrightarrow\sqrt{2x}=2\\ \Leftrightarrow x=2\)
b)
\(3.2.\sqrt{3x}-2.3.\sqrt{3x}+4.\sqrt{3x}=8\\ \Leftrightarrow4\sqrt{3x}=8\\ \Leftrightarrow\sqrt{3x}=2\\\Leftrightarrow x=\dfrac{2^2}{3}=\dfrac{4}{3} \)
c)
\(\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\)
pt<=>\(\sqrt{\left(x+6\right)^3}+\sqrt{x+6}=\left(x^2+4x\right)^3+x^2+4x\)
đặt\(\sqrt{x+6}=a;x^2+4x=b\)
\(a,\) Sửa đề: \(\sqrt{3x^2-12x+16}+\sqrt{y^2-4y+13}=5\)
Ta thấy \(3x^2-12x+16=3\left(x-2\right)^2+4\ge4\Leftrightarrow\sqrt{3x^2-12x+16}\ge\sqrt{4}=2\)
\(y^2-4y+13=\left(y-2\right)^2+9\ge9\Leftrightarrow\sqrt{y^2-4y+13}\ge\sqrt{9}=3\)
Cộng vế theo vế 2 BĐT trên:
\(\sqrt{3x^2-12x+16}+\sqrt{y^2-4y+13}\ge2+3=5\)
Dấu \("="\Leftrightarrow x=y=2\)
Vậy pt có nghiệm \(\left(x;y\right)=\left(2;2\right)\)
\(b,x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\\ \Leftrightarrow x+y+z+4-2\sqrt{x-2}-4\sqrt{y-3}-6\sqrt{z-5}=0\\ \Leftrightarrow\left(x-2-2\sqrt{x-2}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-5+6\sqrt{z-5}+9\right)=0\\ \Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}-1=0\\\sqrt{y-3}-2=0\\\sqrt{z-5}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\y-3=4\\z-5=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\\z=14\end{matrix}\right.\)
Dễ thấy \(x>0\)
Ta có:
\(\left\{{}\begin{matrix}3\sqrt[3]{108x^3+12x}=3\sqrt[3]{2.6x.\left(9x^2+1\right)}\le9x^2+6x+3\\81x^4+5=81x^4+1+4\ge18x^2+4\end{matrix}\right.\)
\(\Rightarrow18x^2+4\le9x^2+6x+3\)
\(\Leftrightarrow9x^2-6x+1\le0\)
\(\Leftrightarrow\left(3x-1\right)^2\le0\)
Dấu = xảy ra khi \(x=\dfrac{1}{3}\)