Giải pt 1) 2-\(\sqrt{\dfrac{x+2}{x-3}}=\sqrt{x+7}\)
2)tìm m để pt \(\dfrac{x-1}{x+1}-2\sqrt{\dfrac{x-1}{x+1}-3m-2=0}\) có nghiệm
Mk đang mắc ở chỗ đặt bằng t rồi chuyển đk của x về điều kiện của t
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Ta có A=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\) với x≥ 9, x ∈ R
Để A > 0 \(\Leftrightarrow\) \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\) > 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-2>0\\\sqrt{x}+1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-2< 0\\\sqrt{x}+1>0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>2\\\sqrt{x}< -1\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 2\\\sqrt{x}>-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>4\\x< 1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 4\\x>1\end{matrix}\right.\end{matrix}\right.\)
Kết hợp với ĐKXĐ\(\Rightarrow\) x ∈ ∅
ĐKXĐ: x≥9, x∈R
Ta có:
A= \(\left[\dfrac{1+\sqrt{x}-\sqrt{x}}{1+\sqrt{x}}\right]\):\(\left[\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{x-2\sqrt{x}-3\sqrt{x}+6}\right]\)
= \(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
=\(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
=\(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
=\(\dfrac{1}{1+\sqrt{x}}\):\(\dfrac{1}{\sqrt{x}-2}\)
=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
Lời giải:
$mA=\sqrt{x}-2$
$\Leftrightarrow \frac{m(2\sqrt{x}-1)}{\sqrt{x}+1}=\sqrt{x}-2$
$\Rightarrow m(2\sqrt{x}-1)=(\sqrt{x}+1)(\sqrt{x}-2)$
$\Leftrightarrow 2m\sqrt{x}-m=x-\sqrt{x}-2$
$\Leftrightarrow x-\sqrt{x}(2m+1)+(m-2)=0(*)$
Để pt ban đầu có 2 nghiệm pb thì $(*)$ phải có 2 nghiệm dương phân biệt.
Điều này xảy ra khi mà:
\(\left\{\begin{matrix}\
\Delta=(2m+1)^2-4(m-2)>0\\
S=2m+1>0\\
P=m-2>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
4m^2+9>0\\
m> \frac{-1}{2}\\
m>2\end{matrix}\right.\Leftrightarrow m>2\)
a) đk:\(\left\{{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-1\ne0\\\sqrt{x}+1\ne0\end{matrix}\right.\\x-1\ne0\\\Rightarrow x\ne1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}y=\sqrt{x}\Rightarrow y>0;y\ne1\\A=\dfrac{\left(y+1\right)^{^2}-\left(y-1\right)^2}{y^2-1}-\dfrac{3y+1}{y^2-1}\end{matrix}\right.\)
\(A=\dfrac{+4y-\left(3y+1\right)}{y^2-1}=\dfrac{y-1}{\left(y-1\right)\left(y+1\right)}=\dfrac{1}{y+1}=\dfrac{1}{\sqrt{x}+1}\)
\(A\left(3-\sqrt{2}\right)=\dfrac{1}{\sqrt{3-\sqrt{2}}+1}\)
c) \(\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{2}\Rightarrow\sqrt{x}=1\Rightarrow x=1\left(loai\right)\) vo nghiệm
d) \(\dfrac{1}{\sqrt{x}+1}< \dfrac{3}{4}\Rightarrow3\sqrt{x}>1\Rightarrow\sqrt{x}>\dfrac{1}{3}\Rightarrow x>\dfrac{1}{9}\)
a) * Đk: \(x\ne\pm1\)
* \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}-1}{x-1} \)
\(A=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)-\left(3\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{x+\sqrt{x}+\sqrt{x}+1-x+\sqrt{x}+\sqrt{x}-1-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{1}{\sqrt{x}-1}\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
Thay \(x=6-2\sqrt{5}\) vào A, ta được:
\(A=\dfrac{\sqrt{5}-1-1}{\sqrt{5}-1+1}=\dfrac{\sqrt{5}-2}{\sqrt{5}}=\dfrac{5-2\sqrt{5}}{5}\)
b: Để \(A< \dfrac{1}{2}\) thì \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{1}{2}< 0\)
\(\Leftrightarrow2\sqrt{x}-2-\sqrt{x}-1< 0\)
\(\Leftrightarrow x< 9\)
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)
2. \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}=\)
\(\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{4+\sqrt{25}}=\sqrt{4+5}=3\)
3. Ta có: VT=\(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}:\sqrt{a}\right).\left(\dfrac{1-\sqrt{a}}{1-a}\right)=\left[\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}.\dfrac{1}{\sqrt{a}}\right].\left[\dfrac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right]=\dfrac{1+\sqrt{a}+a}{\sqrt{a}}.\dfrac{1}{1+\sqrt{a}}=\dfrac{1+\sqrt{a}+a}{\sqrt{a}+a}=\dfrac{1}{\sqrt{a}+a}+1\)
??? Sao rút gọn rồi ra kì vậy nhờ =="
1,
a.
\(\left[{}\begin{matrix}x-5\sqrt{x}+6\ne0\\\sqrt{x}-2\ne0\\3-\sqrt{x}\ne0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)\ne0\\\sqrt{x}\ne2\\\sqrt{x}\ne3\end{matrix}\right.\)
\(\left[{}\begin{matrix}\sqrt{x}\ne3\\\sqrt{x}\ne2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ne9\\x\ne4\end{matrix}\right.\)
Vậy ĐKXĐ : \(\left[{}\begin{matrix}x\ne9\\x\ne4\end{matrix}\right.\)
a) \(ĐK:x\ge0,x\ne1\)
\(=\dfrac{3x+3\sqrt{x}-3-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3x+3\sqrt{x}-3-x+4+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{2x+4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{2\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
b) \(P=\dfrac{2\sqrt{x}}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< 1\)
Kết hợp với đk:
\(\Rightarrow0\le x< 1\)
1) \(\Leftrightarrow4-4\sqrt{\dfrac{x+2}{x-3}}=x+7\)
\(\Leftrightarrow-4\sqrt{\dfrac{x+2}{x-3}}=x+3\)
\(\Leftrightarrow16\dfrac{x+2}{x-3}=x^2+6x+9\)
\(\Leftrightarrow16x+3=x^3+6x^2+9x-3x^2-18x-27\)
\(\Leftrightarrow x^3+3x^2-25x-59=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4,79\\x=-2,2\\x=-5,58\end{matrix}\right.\)
Vậy tập nghiệm....
-Nếu c1 bạn bình phương hai vế thì vế trái là HĐT vẫn thiếu B^2
-Bạn chưa đặt đk gì lsao tương đương như thế được