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12 tháng 11 2018
tìm a nguyên biết (a^2-1)(a^2-4)(a^2-7)(a^2-10)<0
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 5 2023
\(\dfrac{2\sqrt{a}}{\sqrt{a}+3}+\dfrac{\sqrt{a}+1}{\sqrt{a}-3}+\dfrac{3+7\sqrt{a}}{9-a}\)
\(=\dfrac{2\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}+\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}-\dfrac{3+7\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{2a-6\sqrt{a}+a+4\sqrt{a}+3-3-7\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{3a-9\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}=\dfrac{3\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{3\sqrt{a}}{\sqrt{a}+3}\)
23 tháng 8 2021
a, ĐK :a >= 3
\(25\sqrt{\frac{a-3}{25}}-7\sqrt{\frac{4a-12}{9}}-7\sqrt{a^2-9}+18\sqrt{\frac{9a^2-81}{81}}=0\)
\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}\sqrt{a-3}-7\sqrt{\left(a-3\right)\left(a+3\right)}+6\sqrt{\left(a-3\right)\left(a+3\right)}=0\)
\(\Leftrightarrow\sqrt{a-3}\left(5-\frac{14}{3}-\sqrt{a+3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{a-3}=0\\\sqrt{a+3}=\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-\frac{2}{9}\left(loai\right)\end{cases}}\)
b, \(ĐK:x\ge-\frac{1}{2}\)
\(\Leftrightarrow3\sqrt{2x+1}-2\sqrt{2x+1}+\frac{1}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow\frac{4}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow\sqrt{2x+1}=3\)
\(\Leftrightarrow x=4\left(tm\right)\)
23 tháng 8 2021
a) đk: \(a\ge3\)
pt \(\Leftrightarrow25\frac{\sqrt{a-3}}{\sqrt{25}}-7\frac{\sqrt{4\left(a-3\right)}}{\sqrt{9}}-7\sqrt{a^2-9}+18\frac{\sqrt{9\left(a^2-9\right)}}{\sqrt{81}}=0\)
\(\Leftrightarrow5\sqrt{a-3}-\frac{7.2}{3}\sqrt{a-3}-7\sqrt{a^2-9}+\frac{18.3}{9}\sqrt{a^2-9}=0\)
\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}\sqrt{a-3}-7\sqrt{a^2-9}+6\sqrt{a^2-9}=0\)
\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}-\sqrt{a^2-9}=0\)
\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}=\sqrt{a^2-9}\)
\(\Leftrightarrow\frac{1}{9}\left(a-3\right)=a^2-9\)
\(\Leftrightarrow a^2-\frac{1}{9}a-\frac{26}{3}=0\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-\frac{26}{9}\left(loại\right)\end{cases}}\)
2T
A= \(\dfrac{2\sqrt{a}}{\sqrt{a}+3}\)+\(\dfrac{\sqrt{a}+1}{\sqrt{a}-3}\)+\(\dfrac{3+7\sqrt{a}}{9-a}\)
1
H9
HT.Phong (9A5)
CTVHS
14 tháng 8 2023
\(A=\dfrac{2\sqrt{a}}{\sqrt{a}+3}+\dfrac{\sqrt{a}+1}{\sqrt{a}-3}+\dfrac{3-7\sqrt{a}}{9-a}\) (ĐK: \(x\ge0,x\ne9\))
\(A=\dfrac{2\sqrt{a}}{\sqrt{a}+3}+\dfrac{\sqrt{a}+1}{\sqrt{a}-3}-\dfrac{3+7\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(A=\dfrac{2\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}+\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}-\dfrac{3+7\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)
\(A=\dfrac{2a-6\sqrt{a}+a+\sqrt{a}+3\sqrt{a}+3-3-7\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(A=\dfrac{3a-9\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(A=\dfrac{3\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(A=\dfrac{3\sqrt{a}}{\sqrt{a}+3}\)
S
31 tháng 8 2019
\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\frac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\frac{6+2\sqrt{5}}{4}=\frac{32-6-2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}=\frac{14-\sqrt{5}}{2}\) \(\left(\frac{9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2-\left(\frac{9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2=\left(\frac{9-2\sqrt{14}-9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)\left(\frac{9-2\sqrt{14}+9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)=\frac{-72\sqrt{14}}{\sqrt{7}-\sqrt{2}}\)
8 tháng 7 2019
ĐKXĐ: \(a\ge0,a\ne9\)
a)\(P=\frac{2\sqrt{a}}{\sqrt{a}+3}+\frac{\sqrt{a}+1}{\sqrt{a}-3}+\frac{3+7\sqrt{a}}{9-a}.\)
\(=\frac{2\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}+\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}-\frac{3+7\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(=\frac{2\sqrt{a}\left(\sqrt{a}-3\right)+\left(\sqrt{a}+1\right)\left(\sqrt{a}+3\right)-\left(3+7\sqrt{a}\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(=\frac{2a-6\sqrt{a}+a+3\sqrt{a}+\sqrt{a}+3-3-7\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(=\frac{3a-9\sqrt{a}}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}=\frac{3\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(=\frac{3\sqrt{a}}{\sqrt{a}+3}\)
b) Để P<1 hay \(\frac{3\sqrt{a}}{\sqrt{a}+3}< 1\Leftrightarrow3\sqrt{a}< \sqrt{a}+3\Leftrightarrow\sqrt{a}< \frac{3}{2}\Leftrightarrow0\le a< \frac{9}{4}\)(thỏa mãn ĐKXĐ)
Vậy với \(0\le a< \frac{9}{4}\)thì P<1.
(p/s đừng ti ck cho câu trả lời này)
5 tháng 8 2022
1: \(=\left(a-3\right)\cdot\dfrac{\left|b\right|}{a-3}=\left|b\right|\)
2: \(\dfrac{1}{3+a}\cdot\sqrt{\dfrac{a^2+6a+9}{b^2}}\)
\(=\dfrac{1}{a+3}\cdot\dfrac{\left|a+3\right|}{b}=\pm\dfrac{1}{b}\)
3: \(=\left|a+1\right|-\dfrac{3a}{a-2}\cdot\dfrac{\left|a-2\right|}{3}\)
\(=\left|a+1\right|-a\)
4: \(=-6\sqrt{3}+6+28+6\sqrt{3}=34\)
4 tháng 12 2023
a: Khi x=16 thì \(A=\dfrac{2\cdot\sqrt{16}}{\sqrt{16}+3}=\dfrac{2\cdot4}{4+3}=\dfrac{8}{7}\)
b: P=A+B
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{7\sqrt{x}+3}{9-x}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{7\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+7\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+7\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3x+5\sqrt{x}+6}{x-9}\)
1 tháng 1 2022
a: Thay x=16 vào A, ta được:
\(A=\dfrac{2\cdot4}{4+3}=\dfrac{8}{7}\)