Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
ĐK:x>=1
\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)
\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)
\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(ĐK:x\ge3\)
\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)
\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}=0\) (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
<=> x + 1 = 16
<=> x = 15 (nhận)
~ ~ ~
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
<=> x + 5 = 4
<=> x = - 1 (nhận)
a) Thay x=64 vào Q ta có:
\(Q=\dfrac{\sqrt{64}-2}{\sqrt{64}-3}=\dfrac{8-2}{8-3}=\dfrac{6}{5}\)
b) \(P=\dfrac{x}{x-4}-\dfrac{1}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\)
\(P=\dfrac{x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{\sqrt{x}}{\sqrt{x}-2}\left(dpcm\right)\)
E = \(\dfrac{x+2\sqrt{x}+1}{\sqrt{x}+1}+\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\) = \(\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
E = \(\sqrt{x}+1+\sqrt{x}\) = \(2\sqrt{x}+1\)
F = \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}+1}{3-\sqrt{x}}-\dfrac{3-11\sqrt{x}}{x-9}\)
F = \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
F = \(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-\left(3-11\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
F = \(\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}+\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
F = \(\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\) = \(\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\) = \(\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
G = \(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{4\sqrt{x}-4}{4-x}\)
G = \(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{4\sqrt{x}-4}{x-4}\)
G = \(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
G = \(\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(4\sqrt{x}-4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
G = \(\dfrac{x+2\sqrt{x}+3\sqrt{x}+6-\left(x-2\sqrt{x}-\sqrt{x}+2\right)-\left(4\sqrt{x}-4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
G = \(\dfrac{x+5\sqrt{x}+6-x+2\sqrt{x}+\sqrt{x}-2-4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
G = \(\dfrac{4\sqrt{x}+8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\) = \(\dfrac{4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\) = \(\dfrac{4}{\sqrt{x}-2}\)
4 , Ta có :
\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x-9}{x-9}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}-\dfrac{3\left(x-3\right)}{x-9}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x+9}{x-9}\)
\(=\dfrac{3\sqrt{x}+9}{x-9}\)
\(=\dfrac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3}{\sqrt{x}-3}\)
2 , Ta có :
\(\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}+1}=\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{\left(x-1\right)\left(\sqrt{x}-1\right)}{x-1}\)
\(=\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x\sqrt{x}-x-\sqrt{x}+1}{x-1}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}+x+\sqrt{x}-1}{x-1}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
a) đề sai chút nha :
đặc \(a=\sqrt{x-5}\)
b) đặc điều kiện rồi đưa nó về phương trình hệ quả bằng cách bình phương 2 quế giải bt
c) triệt tiêu bớt trên tử và mẩu \(\Rightarrow pt\Leftrightarrow1-x=\dfrac{3}{4}\)
a) điều kiện xác định : \(x\ge1\)
ta có : \(\sqrt{\dfrac{x-1}{4}}-3=\sqrt{\dfrac{4x-4}{9}}\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-3=\dfrac{2}{3}\sqrt{x-1}\)
\(\Leftrightarrow\dfrac{1}{6}\sqrt{x-1}=-3\left(vôlí\right)\) vậy phương trình vô nghiệm
b) điều kiện xác định \(x\ge3\)
ta có : \(\sqrt{x^2-4x+4}+\sqrt{x^2+6x+9}=x-3\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}+\sqrt{\left(x+3\right)^2}=x-3\) \(\Leftrightarrow\left|x-2\right|+\left|x+3\right|=x-3\)
\(\Leftrightarrow x-2+x+3=x-3\Leftrightarrow x=-4\left(L\right)\) vậy phương trình vô nghiệm
c) điều kiện xác định : \(\left[{}\begin{matrix}x\ge\dfrac{3}{2}\\x< 1\end{matrix}\right.\)
ta có : \(\sqrt{\dfrac{2x-3}{x-1}}=2\) \(\Leftrightarrow\dfrac{2x-3}{x-1}=4\Leftrightarrow2x-3=4x-4\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tmđk\right)\) vậy \(x=\dfrac{1}{2}\)
a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)
\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)
b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)
\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)
c: \(C=x-4+\left|x-4\right|\)
=x-4+x-4
=2x-8
\(Q=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)+9\sqrt{x}-4+\left(4x-4\sqrt{x}\right)\left(\sqrt{x}+4\right)}{x-16}\)
\(=\dfrac{x+4\sqrt{x}+4x\sqrt{x}+16x-4x-16\sqrt{x}}{x-16}\)
\(=\dfrac{13x+4x\sqrt{x}-12\sqrt{x}}{x-16}\)