c1: Rút gọn biểu thức A=\(\left(\dfrac{1}{x-2\sqrt{x}}-\dfrac{2}{6-3\sqrt{x}}\right):\left(\dfrac{2}{3}+\dfrac{1}{\sqrt{x}}\right)\)
c2: Cho phương trình: \(x^2-2\left(2m-1\right)x+m^2-4m=0\left(1\right)\)
Tìm m để phương trình (1) có hai nghiệm phân biệt x1, x2 thoả mãn hệ thức \(x_1+x_2=\dfrac{-8}{x_1+x_2}\)
1:
\(=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{2}{3\sqrt{x}-6}\right):\dfrac{2\sqrt{x}+3}{3\sqrt{x}}\)
\(=\dfrac{3+2\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{3\sqrt{x}}{2\sqrt{x}+3}=\dfrac{1}{\sqrt{x}-2}\)
Giải PT :
\(\dfrac{13\left(1-2x^2\right)}{\sqrt{1-x^2}}+\dfrac{9\left(1+2x^2\right)}{\sqrt{1+x^2}}=0\)
\(ĐK:-1\le x\le1\\ PT\Leftrightarrow13\left(1-2x^2\right)\sqrt{\left(1-x^2\right)\left(1+x^2\right)}+9\left(1+2x^2\right)\sqrt{\left(1+x^2\right)\left(1-x^2\right)}=0\\ \Leftrightarrow\sqrt{1-x^4}\left(13-26x^2+9+18x^2\right)=0\\ \Leftrightarrow\sqrt{1-x^4}\left(22-8x^2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}1-x^4=0\\22-8x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(1+x^2\right)\left(1-x\right)\left(1+x\right)=0\\x^2=\dfrac{22}{8}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\\\left[{}\begin{matrix}x=\dfrac{\sqrt{11}}{2}\left(ktm\right)\\x=-\dfrac{\sqrt{11}}{2}\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
\(D=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-1\right)\)
1, Rút gọn D
2, Tìm x để \(D=\left(4-\dfrac{x-\sqrt{x}+13}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+1}{2\sqrt{x}+1}\)
1: \(D=\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
2: \(\Leftrightarrow D=\dfrac{4\sqrt{x}+12-x+\sqrt{x}-13}{\sqrt{x}+3}\cdot\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-x+5\sqrt{x}-1}{\sqrt{x}+3}\cdot\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)
\(\Leftrightarrow\dfrac{-x+5\sqrt{x}-1}{\sqrt{x}+3}\cdot\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\cdot\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}=1\)
\(\Leftrightarrow\left(-x+5\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)=\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+3\right)\)
\(\Leftrightarrow-2x\sqrt{x}-x+10x+5\sqrt{x}-2\sqrt{x}-1=x\sqrt{x}+3x+x+3\sqrt{x}+\sqrt{x}+3\)
\(\Leftrightarrow-2x\sqrt{x}+9x-3\sqrt{x}-1=x\sqrt{x}+4x+4\sqrt{x}+3\)
\(\Leftrightarrow-3x\sqrt{x}+5x-7\sqrt{x}-4=0\)
Bạn xem lại đề nhé, nghiệm rất xấu
Giải các bất phương trình, hệ phương trình
a) \(\dfrac{x^2\left(3x-2\right)\left(x^2-1\right)}{\left(-x^2+2x-3\right)\left(2-x\right)^2}\ge0\)
b) \(\dfrac{x-5}{x-1}>2\)
c) \(2x-\sqrt{x^2-5x-14}< 1\)
d) \(x+\sqrt{x^2-4x-5}< 4\)
e) \(\left\{{}\begin{matrix}\left(4-x\right)\left(x^2-2x-3\right)< 0\\x^2\ge\left(x^2-x-3\right)^2\end{matrix}\right.\)
Giải các bất phương trình sau:
\(a,\left(x+1\right)\left(x+4\right)< 5\sqrt{x^2+5x+28}\)
\(b,4\sqrt{x}+\dfrac{2}{\sqrt{x}}< 2x+\dfrac{1}{2x}+2\)
a, ĐKXĐ : \(D=R\)
BPT \(\Leftrightarrow x^2+5x+4< 5\sqrt{x^2+5x+4+24}\)
Đặt \(x^2+5x+4=a\left(a\ge-\dfrac{9}{4}\right)\)
BPTTT : \(5\sqrt{a+24}>a\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a+24\ge0\\a< 0\end{matrix}\right.\\\left\{{}\begin{matrix}a\ge0\\25\left(a+24\right)>a^2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-24\le a< 0\\\left\{{}\begin{matrix}a^2-25a-600< 0\\a\ge0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-24\le a< 0\\0\le a< 40\end{matrix}\right.\)
\(\Leftrightarrow-24\le a< 40\)
- Thay lại a vào ta được : \(\left\{{}\begin{matrix}x^2+5x-36< 0\\x^2+5x+28\ge0\end{matrix}\right.\)
\(\Leftrightarrow-9< x< 4\)
Vậy ....
b, ĐKXĐ : \(x>0\)
BĐT \(\Leftrightarrow2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< x+\dfrac{1}{4x}+1\)
- Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a\left(a\ge\sqrt{2}\right)\)
\(\Leftrightarrow a^2=x+\dfrac{1}{4x}+1\)
BPTTT : \(2a\le a^2\)
\(\Leftrightarrow\left[{}\begin{matrix}a\le0\\a\ge2\end{matrix}\right.\)
\(\Leftrightarrow a\ge2\)
\(\Leftrightarrow a^2\ge4\)
- Thay a vào lại BPT ta được : \(x+\dfrac{1}{4x}-3\ge0\)
\(\Leftrightarrow4x^2-12x+1\ge0\)
\(\Leftrightarrow x=(0;\dfrac{3-2\sqrt{2}}{2}]\cup[\dfrac{3+2\sqrt{2}}{2};+\infty)\)
Vậy ...
Tập nghiệm của bất pt
a) \(\left|3x+1\right|>2\)
b) \(\left|2x-1\right|\le1\)
c) \(\left|\dfrac{2}{x-13}\right|>\dfrac{8}{9}\). Số nghiệm nguyên nhỏ hơn 13 của bất pt
d) \(\dfrac{\left|x+2\right|-x}{x}\le2\)
a, \(\left|3x+1\right|>2\)
\(\Leftrightarrow\left(\left|3x+1\right|\right)^2>4\)
\(\Leftrightarrow9x^2+6x+1>4\)
\(\Leftrightarrow9x^2+6x-3>0\)
\(\Leftrightarrow3\left(3x-1\right)\left(x+1\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{3}\\x< -1\end{matrix}\right.\)
b, \(\left|2x-1\right|\le1\)
\(\Leftrightarrow\left(\left|2x-1\right|\right)^2\le1\)
\(\Leftrightarrow4x^2-4x+1\le1\)
\(\Leftrightarrow4x\left(x-1\right)\le0\)
\(\Leftrightarrow0\le x\le1\)
c, ĐK: \(x\ne13\)
\(\left|\dfrac{2}{x-13}\right|>\dfrac{8}{9}\)
\(\Leftrightarrow\dfrac{1}{\left|x-13\right|}>\dfrac{4}{9}\)
\(\Leftrightarrow4\left|x-13\right|< 9\)
\(\Leftrightarrow16\left(x^2-26x+169\right)< 81\)
\(\Leftrightarrow16x^2-416x+2623< 0\)
\(\Leftrightarrow\dfrac{43}{4}< x< \dfrac{61}{4}\)
\(\Rightarrow\) Có hai giả trị thỏa mãn yêu cầu bài toán
Giải hệ bằng phương pháp phân tích nhân tử
a) \(\left\{{}\begin{matrix}\dfrac{1}{x^2}+\dfrac{1}{y^2}=1\\\sqrt{x^2-1}+\sqrt{y^2-1}=\sqrt{xy+2}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\left(x-y\right)\left(x^2+y^2\right)=13\\\left(x+y\right)\left(x^2-y^2\right)=25\end{matrix}\right.\)
Giải các bất phương trình sau:
a) \(\dfrac{2x^2}{\left(3-\sqrt{9+2x}\right)^2}< x+21\)
b) \(\sqrt{x-\dfrac{1}{x}}+\sqrt{1-\dfrac{1}{x}}\ge x\)
giải bất phương trình \(\left(\sqrt{13}-\sqrt{2x^2-2x+5}-\sqrt{2x^2-4x+4}\right)\left(x^6-x^3+x^2-x+1\right)\ge0\)
Do \(x^6-x^3+x^2-x+1=\left(x^3-\dfrac{1}{2}\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}>0\) ; \(\forall x\) nên BPT tương đương:
\(\sqrt{13}-\sqrt{2x^2-2x+5}-\sqrt{2x^2-4x+4}\ge0\)
\(\Leftrightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}\le\sqrt{26}\) (1)
Ta có:
\(VT=\sqrt{\left(2x-1\right)^2+3^2}+\sqrt{\left(2-2x\right)^2+2^2}\ge\sqrt{\left(2x-1+2-2x\right)^2+\left(3+2\right)^2}=\sqrt{26}\) (2)
\(\Rightarrow\left(1\right);\left(2\right)\Rightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}=\sqrt{26}\)
Dấu "=" xảy ra khi và chỉ khi \(2\left(2x-1\right)=3\left(2-2x\right)\Leftrightarrow x=\dfrac{4}{5}\)
Vậy BPT có nghiệm duy nhất \(x=\dfrac{4}{5}\)
1)\(\sqrt{x+3}\) > 2
2) \(\dfrac{1+\sqrt{x}}{\sqrt{x}-2}\)<1
3) \(\left(\sqrt{x}-1\right)\).\(\left(\sqrt{x}-3\right)\)-5=\(\sqrt{x}\) \(\left(\sqrt{x}+2\right)-5\)
tìm x mn giúp mình nha plsss
1: ĐKXĐ: x+3>=0
=>x>=-3
\(\sqrt{x+3}>2\)
=>x+3>4
=>x>4-3=1
2: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >4\end{matrix}\right.\)
\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}< 1\)
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-1< 0\)
=>\(\dfrac{\sqrt{x}+1-\sqrt{x}+2}{\sqrt{x}-2}< 0\)
=>\(\dfrac{3}{\sqrt{x}-2}< 0\)
=>\(\sqrt{x}-2< 0\)
=>\(\sqrt{x}< 2\)
=>0<=x<4
3: ĐKXĐ: x>=0
\(\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)-5=\sqrt{x}\left(\sqrt{x}+2\right)-5\)
=>\(x-4\sqrt{x}+3-5=x+2\sqrt{x}-5\)
=>\(x-4\sqrt{x}-2-x-2\sqrt{x}+5=0\)
=>\(-6\sqrt{x}+3=0\)
=>\(-6\sqrt{x}=-3\)
=>\(\sqrt{x}=\dfrac{1}{2}\)
=>x=1/4(nhận)
