rút gọn biểu thức
A=\(4\sqrt{\dfrac{25x}{4}}-\dfrac{8}{3}\sqrt{\dfrac{9x}{4}}-\dfrac{4}{3x}\sqrt{\dfrac{9x^3}{64}}\) với \(x>0\)
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Những câu hỏi liên quan
5 tháng 7 2021
a) Ta có: \(A=3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+30\)
\(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+30\)
\(=14\sqrt{2x}+30\)
b) Ta có: \(B=4\sqrt{\dfrac{25x}{4}}-\dfrac{8}{3}\sqrt{\dfrac{9x}{4}}-\dfrac{4}{3x}\cdot\sqrt{\dfrac{9x^3}{64}}\)
\(=4\cdot\dfrac{5\sqrt{x}}{2}-\dfrac{8}{3}\cdot\dfrac{3\sqrt{x}}{2}-\dfrac{4}{3x}\cdot\dfrac{3x\sqrt{x}}{8}\)
\(=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}\sqrt{x}\)
\(=\dfrac{11}{2}\sqrt{x}\)
c) Ta có: \(\dfrac{y}{2}+\dfrac{3}{4}\sqrt{9y^2-6y+1}-\dfrac{3}{2}\)
\(=\dfrac{1}{2}y+\dfrac{3}{4}\left(1-3y\right)-\dfrac{3}{2}\)
\(=\dfrac{1}{2}y+\dfrac{3}{4}-\dfrac{9}{4}y-\dfrac{3}{2}\)
\(=-\dfrac{7}{4}y-\dfrac{3}{4}\)
15 tháng 10 2023
a: ĐKXĐ: x-5>=0
=>x>=5
\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)
=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
=>\(2\sqrt{x-5}=4\)
=>x-5=4
=>x=9(nhận)
b: ĐKXĐ: x-1>=0
=>x>=1
\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}=4\)
=>\(\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=4\)
=>\(-2\sqrt{x-1}=4\)
=>\(\sqrt{x-1}=-2\)(vô lý)
Vậy: Phương trình vô nghiệm
c: ĐKXĐ: x-2>=0
=>x>=2
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot\sqrt{9x-18}+6\cdot\sqrt{\dfrac{x-2}{81}}=-4\)
=>\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)
=>\(\sqrt{x-2}\left(\dfrac{1}{3}-2+\dfrac{2}{3}\right)=-4\)
=>\(-\sqrt{x-2}=-4\)
=>x-2=16
=>x=18(nhận)
d: ĐKXĐ: x+3>=0
=>x>=-3
\(\sqrt{9x+27}+4\sqrt{x+3}-\dfrac{3}{4}\cdot\sqrt{16x+48}=0\)
=>\(3\sqrt{x+3}+4\sqrt{x+3}-\dfrac{3}{4}\cdot4\sqrt{x+3}=0\)
=>\(4\sqrt{x+3}=0\)
=>x+3=0
=>x=-3(nhận)
NV
15 tháng 10 2023
a) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
= \(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\)
= \(2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
= \(2\sqrt{x-5}=4\)
= \(\sqrt{x-5}=2\)
= \(\left|x-5\right|=4\)
=> \(x-5=\pm4\)
\(x=\pm4+5\)
\(x=9;x=1\)
Vậy x=9; x=1
2 tháng 9 2021
\(a.\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)( x lớn hơn hoặc =1)
\(\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}\)+2=0
\(\sqrt{x-1}\left(1+\sqrt{4}-\sqrt{25}\right)=-2\)
\(\sqrt{x-1}\left(1+2-5\right)=-2\)
\(\sqrt{x-1}.\left(-2\right)=-2\)
\(\sqrt{x-1}=-2.2\)
\(\sqrt{x-1}-4\)(ko thỏa mãn)
b)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9\left(x-1\right)}+24\dfrac{\sqrt{x-1}}{8}=-17\)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}.3\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\left(\dfrac{1}{2}-\dfrac{9}{2}+3\right)\sqrt{x-1}=-17\)
\(7\sqrt{x-1}=-17\)
\(\sqrt{x-1}=-\dfrac{17}{7}\)(ko thỏa mãn căn bậc 2 ko có số âm)
2 tháng 9 2021
a: Ta có: \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=-2\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow x-1=1\)
hay x=2
5 tháng 9 2023
a: ĐKXĐ: x>=3
Sửa đề: \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}}-3=0\)
=>\(2\sqrt{x-3}-3\sqrt{x-3}+\dfrac{5}{2}\sqrt{x-3}-3=0\)
=>\(\dfrac{3}{2}\sqrt{x-3}=3\)
=>\(\sqrt{x-3}=2\)
=>x-3=4
=>x=7(nhận)
b: ĐKXĐ: x>=0
\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< =-\dfrac{3}{4}\)
=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{3}{4}< =0\)
=>\(\dfrac{4\sqrt{x}-8+3\sqrt{x}+3}{4\left(\sqrt{x}+1\right)}< =0\)
=>\(7\sqrt{x}-5< =0\)
=>\(\sqrt{x}< =\dfrac{5}{7}\)
=>0<=x<=25/49
c: ĐKXĐ: x>=5
\(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)
=>\(3\sqrt{x-5}-14\cdot\dfrac{\sqrt{x-5}}{7}+\dfrac{1}{4}\cdot2\cdot\sqrt{x-5}=3\)
=>\(\dfrac{3}{2}\sqrt{x-5}=3\)
=>\(\sqrt{x-5}=2\)
=>x-5=4
=>x=9(nhận)
11 tháng 7 2021
Bài 1:
a) \(\dfrac{a+\sqrt{a}}{\sqrt{a}}=\sqrt{a}+1\)
b) \(\dfrac{\sqrt{\left(x-3\right)^2}}{3-x}=\dfrac{\left|x-3\right|}{3-x}=\pm1\)
Bài 2:
a) \(\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}=\pm\dfrac{1}{3x+1}\)
b) \(4-x-\sqrt{x^2-4x+4}=4-x-\left|x-2\right|=\left[{}\begin{matrix}6-2x\left(x\ge2\right)\\2\left(x< 2\right)\end{matrix}\right.\)
21 tháng 7 2023
a: \(Q=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
b: Khi x=4+2căn 3 thì \(Q=\dfrac{\sqrt{3}+1-2}{\sqrt{3}+1+2}=\dfrac{-3+2\sqrt{3}}{3}\)
c: Q=3
=>3căn x+6=căn x-2
=>2căn x=-8(loại)
d: Q>1/2
=>Q-1/2>0
=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{1}{2}>0\)
=>2căn x-4-căn x-2>0
=>căn x>6
=>x>36
d: Q nguyên
=>căn x+2-4 chia hết cho căn x+2
=>căn x+2 thuộc Ư(-4)
=>căn x+2 thuộc {2;4}
=>x=0 hoặc x=4(nhận)
7 tháng 10 2021
c: Ta có: \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)
\(\Leftrightarrow2\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=4\)
hay x=5
e: Ta có: \(\sqrt{4x^2-28x+49}-5=0\)
\(\Leftrightarrow\left|2x-7\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=5\\2x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
8 tháng 10 2021
a. ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow \sqrt{(x-2)^2}=2-x$
$\Leftrightarrow |x-2|=2-x$
$\Leftrightarrow 2-x\geq 0$
$\Leftrightarrow x\leq 2$
b. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-2}-\frac{1}{5}\sqrt{25}.\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 2\sqrt{x-2}-\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 1=2\sqrt{x-2}$
$\Leftrightarrow \frac{1}{2}=\sqrt{x-2}$
$\Leftrightarrow \frac{1}{4}=x-2$
$\Leftrightarrow x=\frac{9}{4}$ (tm)
H9
HT.Phong (9A5)
CTVHS
4 tháng 9 2023
\(A=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\) (ĐK: \(x\ge0;x\ne\dfrac{1}{9}\))
\(A=\left[\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}\right)^2-1^2}\right]:\left[\dfrac{\left(3\sqrt{x}+1\right)\cdot1}{3\sqrt{x}+1}-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right]\)
\(A=\left[\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(A=\left[\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}-\dfrac{3\sqrt{x}-1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\dfrac{3}{3\sqrt{x}+1}\)
\(A=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\cdot\dfrac{3\sqrt{x}+1}{3}\)
\(A=\dfrac{3x+3\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\cdot\dfrac{3\sqrt{x}+1}{3}\)
\(A=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\cdot\dfrac{3\sqrt{x}+1}{3}\)
\(A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}-1}\)
\(A=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)
4 tháng 9 2023
\(A=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right):\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\cdot\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1+5\sqrt{x}+1}{3\sqrt{x}-1}\cdot\dfrac{1}{3}\)
\(=\dfrac{3x+3\sqrt{x}}{3\sqrt{x}-1}\cdot\dfrac{1}{3}\)
\(=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)
\(A=4\sqrt{\dfrac{25x}{4}}-\dfrac{8}{3}\sqrt{\dfrac{9x}{4}}-\dfrac{4}{3x}\sqrt{\dfrac{9x^3}{64}}\)
\(A=4\left|\dfrac{5\sqrt{x}}{2}\right|-\dfrac{8}{3}\left|\dfrac{3\sqrt{x}}{2}\right|-\dfrac{4}{3x}\left|\dfrac{3x\sqrt{x}}{8}\right|\)
Vì \(x>0\) nên:
\(A=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}\sqrt{x}=\dfrac{11\sqrt{x}}{2}\)