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22 tháng 6 2017
a) \(\dfrac{12}{x-1}-\dfrac{8}{x+1}=1\) \(\Leftrightarrow\) \(\dfrac{12\left(x+1\right)-8\left(x-1\right)}{x^2-1}=1\)
\(\Leftrightarrow\) \(\dfrac{12x+12-8x+8}{x^2-1}=1\) \(\Leftrightarrow\) \(\dfrac{4x+20}{x^2-1}=1\)
\(\Leftrightarrow\) \(x^2-1=4x+20\) \(\Leftrightarrow\) \(x^2-4x-21=0\)
giải pt ta có 2 nghiệm : \(x_1=7;x_2=-3\)
vậy phương trình có 2 nghiệm \(x=7;x=-3\)
b) \(\dfrac{16}{x-3}+\dfrac{30}{1-x}=3\) \(\Leftrightarrow\) \(\dfrac{16\left(1-x\right)+30\left(x-3\right)}{\left(x-3\right)\left(1-x\right)}=3\)
\(\Leftrightarrow\) \(\dfrac{16-16x+30x-90}{x-x^2-3+3x}=3\) \(\Leftrightarrow\) \(\dfrac{14x-74}{-x^2+4x-3}=3\)
\(\Leftrightarrow\) \(3\left(-x^2+4x-3\right)=14x-74\)
\(\Leftrightarrow\) \(-3x^2+12x-9=14x-74\)
\(\Leftrightarrow\) \(3x^2-2x-65=0\)
giải pt ta có 2 nghiệm : \(x_1=5;x_2=\dfrac{-13}{3}\)
vậy phương trình có 2 nghiệm \(x=5;x=\dfrac{-13}{3}\)
DA
29 tháng 9 2017
Bài 1:
a/ \(\sqrt{\dfrac{2x^2-4x+2}{6}}=1\) .
\(\Leftrightarrow\dfrac{2\left(x^2-2x+1\right)}{6}=1\)
\(\Leftrightarrow\dfrac{\left(x-1\right)^2}{3}=1\)
\(\Leftrightarrow\left(x-1\right)^2=3\) \(\Rightarrow\left[{}\begin{matrix}x-1=\sqrt{3}\\x-1=-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}+1\\x=-\sqrt{3}+1\end{matrix}\right.\)
vậy tập nghiệm của phương trình S=\(\left\{1-\sqrt{3};\sqrt{3}+1\right\}\)
b/ ta có: \(\dfrac{6}{x-4}=\sqrt{2}\Leftrightarrow\sqrt{2}\left(x-4\right)=6\)
\(\Leftrightarrow x\sqrt{2}-4\sqrt{2}=6\)
\(\Leftrightarrow x\sqrt{2}=6+4\sqrt{2}\)
\(\Leftrightarrow x=\dfrac{6+4\sqrt{2}}{2}=4+3\sqrt{2}\)
vậy \(x=4+3\sqrt{2}\) là nghiệm của phương trình
c/ \(\sqrt{\dfrac{20}{2x^2-8x+8}}=\sqrt{5}\)
\(\Leftrightarrow\left(\sqrt{\dfrac{20}{2x^2-8x+8}}\right)^2=\left(\sqrt{5}\right)^2\)
\(\Leftrightarrow\dfrac{20}{2\left(x^2-4x+4\right)}=5\)
\(\Leftrightarrow\dfrac{10}{\left(x-2\right)^2}=\dfrac{10}{2}\)
\(\Rightarrow\left(x-2\right)^2=2\) \(\Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{2}\\x-2=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{2}\\x=2-\sqrt{2}\end{matrix}\right.\)
vậy tập nghiệm của phương trình \(S=\left\{2+\sqrt{2};2-\sqrt{2}\right\}\)
DA
29 tháng 9 2017
Bài 2:
a/ đặt A= \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\)
\(\Leftrightarrow A^2=3+\sqrt{5}+3-\sqrt{5}-2\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)
\(\Leftrightarrow A^2=6-2\sqrt{9-5}\)
\(\Leftrightarrow A^2=6-2\sqrt{4}=6-4=2\)
\(\Rightarrow A=\sqrt{2}\)
\(\Rightarrow\)\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\) = \(\sqrt{2}\)
\(\Rightarrow\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}=\sqrt{2}-\sqrt{2}=0\)
b/ \(\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}\)
\(=\dfrac{\sqrt{12}}{\sqrt{15}}+\dfrac{\sqrt{75}}{\sqrt{15}}+\dfrac{\sqrt{27}}{\sqrt{15}}=\sqrt{\dfrac{12}{15}}+\sqrt{\dfrac{75}{15}}+\sqrt{\dfrac{27}{15}}\)
\(=\dfrac{2\sqrt{5}}{5}+\sqrt{5}+\dfrac{3\sqrt{5}}{5}=\left(\dfrac{2\sqrt{5}}{5}+\dfrac{3\sqrt{5}}{5}\right)+\sqrt{5}\)
\(=\sqrt{5}+\sqrt{5}=2\sqrt{5}\)
c/ \(\left(12\sqrt{20}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
\(=\left(24\sqrt{5}-80\sqrt{2}+105\sqrt{2}\right):\sqrt{10}\)
\(=\left(24\sqrt{5}+25\sqrt{2}\right):\sqrt{10}=\dfrac{24\sqrt{5}}{\sqrt{10}}+\dfrac{25\sqrt{2}}{\sqrt{10}}\)
\(=12\sqrt{2}+5\sqrt{5}\)
22 tháng 6 2017
a) \(\dfrac{12}{x-1}-\dfrac{8}{x+1}=1\) \(\Leftrightarrow\) \(\dfrac{12\left(x+1\right)-8\left(x-1\right)}{x^2-1}=1\)
\(\Leftrightarrow\) \(\dfrac{12x+12-8x+8}{x^2-1}=1\) \(\Leftrightarrow\) \(\dfrac{4x+20}{x^2-1}=1\)
\(\Leftrightarrow\) \(x^2-1=4x+20\) \(\Leftrightarrow\) \(x^2-4x-21=0\)
giải pt ta có 2 nghiệm : \(x_1=7;x_2=-3\)
vậy phương trình có 2 nghiệm \(x=7;x=-3\)
b) \(\dfrac{16}{x-3}+\dfrac{30}{1-x}=3\) \(\Leftrightarrow\) \(\dfrac{16\left(1-x\right)+30\left(x-3\right)}{\left(x-3\right)\left(1-x\right)}=3\)
\(\Leftrightarrow\) \(\dfrac{16-16x+30x-90}{x-x^2-3+3x}=3\) \(\Leftrightarrow\) \(\dfrac{14x-74}{-x^2+4x-3}=3\)
\(\Leftrightarrow\) \(3\left(-x^2+4x-3\right)=14x-74\)
\(\Leftrightarrow\) \(-3x^2+12x-9=14x-74\)
\(\Leftrightarrow\) \(3x^2-2x-65=0\)
giải pt ta có 2 nghiệm : \(x_1=5;x_2=\dfrac{-13}{3}\)
vậy phương trình có 2 nghiệm \(x=5;x=\dfrac{-13}{3}\)
27 tháng 12 2018
c) ĐK: x\(\ne3,x\ne-2\)
\(\dfrac{x^2-3x+5}{\left(x-3\right)\left(x+2\right)}=\dfrac{1}{x-3}\Leftrightarrow\dfrac{x^2-3x+5}{\left(x-3\right)\left(x+2\right)}=\dfrac{x+2}{\left(x-3\right)\left(x+2\right)}\Leftrightarrow x^2-3x+5=x+2\Leftrightarrow x^2-4x+3=0\Leftrightarrow x^2-x-3x+3=0\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=1\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)
Vậy S={1}
d) ĐK: \(x\ne2,x\ne-4\)
\(\dfrac{2x}{x-2}-\dfrac{x}{x+4}=\dfrac{8x+8}{\left(x-2\right)\left(x+4\right)}\Leftrightarrow\dfrac{2x\left(x+4\right)}{\left(x-2\right)\left(x+4\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+4\right)}=\dfrac{8x+8}{\left(x-2\right)\left(x+4\right)}\Leftrightarrow\dfrac{2x^2+8x}{\left(x-2\right)\left(x+4\right)}-\dfrac{x^2-2x}{\left(x-2\right)\left(x+4\right)}=\dfrac{8x+8}{\left(x-2\right)\left(x+4\right)}\Leftrightarrow\dfrac{2x^2+8x-x^2+2x}{\left(x-2\right)\left(x+4\right)}=\dfrac{8x+8}{\left(x-2\right)\left(x+4\right)}\Leftrightarrow x^2+10x=8x+8\Leftrightarrow x^2+2x-8=0\Leftrightarrow x^2-2x+4x-8=0\Leftrightarrow x\left(x-2\right)+4\left(x-2\right)=0\Leftrightarrow\left(x-2\right)\left(x+4\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=2\left(ktm\right)\\x=-4\left(ktm\right)\end{matrix}\right.\)
Vậy phương trình vô nghiệm
12 tháng 5 2018
Đề bài sai: Khi \(x=4\) thì \(A=\dfrac{1}{2};B=\dfrac{28}{9};\dfrac{A}{B}=\dfrac{9}{56};\dfrac{x-2}{4\sqrt{x}}=\dfrac{1}{4}\Rightarrow\dfrac{A}{B}\ne\dfrac{x-2}{4\sqrt{x}}\)
HL
17 tháng 9 2017
\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
\(\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\dfrac{18\left(x+7-x-4\right)}{18\left(x+4\right)\left(x+7\right)}=\dfrac{\left(x+4\right)\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}\)
\(18.3=\left(x+4\right)\left(x+7\right)\)
\(x^2+11x+28-54=0\)
\(x^2+11x-26=0\)
\(\left(x-2\right)\left(x+13\right)=0\)
\(\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)
Theo đề x < 0 nên x = -13
3 tháng 9 2022
1: \(=3\left(x+\dfrac{2}{3}\sqrt{x}+\dfrac{1}{3}\right)\)
\(=3\left(x+2\cdot\sqrt{x}\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{2}{9}\right)\)
\(=3\left(\sqrt{x}+\dfrac{1}{3}\right)^2+\dfrac{2}{3}>=3\cdot\dfrac{1}{9}+\dfrac{2}{3}=1\)
Dấu '=' xảy ra khi x=0
2: \(=x+3\sqrt{x}+\dfrac{9}{4}-\dfrac{21}{4}=\left(\sqrt{x}+\dfrac{3}{2}\right)^2-\dfrac{21}{4}>=-3\)
Dấu '=' xảy ra khi x=0
3: \(A=-2x-3\sqrt{x}+2< =2\)
Dấu '=' xảy ra khi x=0
5: \(=x-2\sqrt{x}+1+1=\left(\sqrt{x}-1\right)^2+1>=1\)
Dấu '=' xảy ra khi x=1
AH
Akai Haruma
Giáo viên
14 tháng 10 2018
1)
ĐK: \(x\geq 5\)
PT \(\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=6\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}=6\Rightarrow \sqrt{x-5}=3\Rightarrow x=3^2+5=14\)
AH
Akai Haruma
Giáo viên
14 tháng 10 2018
2)
ĐK: \(x\geq -1\)
\(\sqrt{x+1}+\sqrt{x+6}=5\)
\(\Leftrightarrow (\sqrt{x+1}-2)+(\sqrt{x+6}-3)=0\)
\(\Leftrightarrow \frac{x+1-2^2}{\sqrt{x+1}+2}+\frac{x+6-3^2}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow (x-3)\left(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}\right)=0\)
Vì \(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}>0, \forall x\geq -1\) nên $x-3=0$
\(\Rightarrow x=3\) (thỏa mãn)
Vậy .............
7 tháng 10 2021
a, \(\frac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}=\frac{\sqrt{10}+\sqrt{6}}{\sqrt{10.3}+\sqrt{6.3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\)
b, Với a;b > 0
\(\frac{a+\sqrt{ab}}{b+\sqrt{ab}}=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{b}+\sqrt{a}\right)}=\frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{ab}}{b}\)
c, Với x >= 0
\(\frac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}=\frac{\left(\sqrt{x}-1\right)\left(4\sqrt{x}+7\right)}{4\sqrt{x}+7}=\sqrt{x}-1\)
d, Với x >= 0 ; x khác 14
\(\frac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)
XO
7 tháng 10 2021
a) \(\frac{\sqrt{10}+\sqrt{6}}{\sqrt{30}+\sqrt{18}}=\frac{\sqrt{10}+\sqrt{6}}{\sqrt{3}\left(\sqrt{10}+\sqrt{6}\right)}=\frac{1}{\sqrt{3}}\)
b) \(\frac{a+\sqrt{ab}}{b+\sqrt{ab}}=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}=\frac{\sqrt{a}}{\sqrt{b}}\)
c) \(\frac{4x+3\sqrt{x}-7}{4\sqrt{x}+7}=\frac{\left(\sqrt{x}-1\right)\left(4\sqrt{x}+7\right)}{\left(4\sqrt{x}+7\right)}=\sqrt{x}-1\)
d) \(\frac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}=\frac{x+\sqrt{x}-4\sqrt{x}-4}{x-4\sqrt{x}+3\sqrt{x}-12}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(\dfrac{450}{x+30}-\dfrac{450}{x}=4\left(đk:x\ne0,x\ne-30\right)\)
\(\Leftrightarrow\dfrac{450x-450x-13500}{x\left(x+30\right)}=4\)
\(\Leftrightarrow\dfrac{-13500}{x\left(x+30\right)}=4\)
\(\Leftrightarrow4x^2+120x=-13500\)
\(\Leftrightarrow4\left(x+15\right)^2=-12600\)
\(\Leftrightarrow\left(x+15\right)^2=-3150\)(vô lý)
Vậy \(x\in\varnothing\)
\(\dfrac{450}{x+30}-\dfrac{450}{x}=4\)
<=> \(\dfrac{450x}{x\left(x+30\right)}-\dfrac{450\left(x+30\right)}{x\left(x+30\right)}=\dfrac{4x\left(x+30\right)}{x\left(x+30\right)}\)
<=> 450x - 450(x + 30) = 4x(x + 30)
<=> 450x - 450x - 480 = 4x2 + 120x
<=> 4x2 - 120x - 480 = 0
<=> x = 33, 57