Giải phương trình:
a) \(2\sqrt{20x}-3\sqrt{45x}+4\sqrt{80x}=33\)
b) \(3\sqrt{9x-27}+2\sqrt{25x-75}-\sqrt{49-147}=48\)
Help me :(((((
Giải phương trình và bất phương trình:
a) \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}-3=0}\)
b) \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\) ≤ \(\dfrac{-3}{4}\)
c) \(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)
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)
Giải các phương trình sau:
a. \(\sqrt{25x+75}+2\sqrt{9x+27}=5\sqrt{x+3}+18\)
b. \(\sqrt{4x-8}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)
a) Ta có: \(\sqrt{25x+75}+2\sqrt{9x+27}=5\sqrt{x+3}+18\)
\(\Leftrightarrow5\sqrt{x+3}+6\sqrt{x+3}-5\sqrt{x+3}=18\)
\(\Leftrightarrow\sqrt{x+3}=3\)
\(\Leftrightarrow x+3=9\)
hay x=6
b) Ta có: \(\sqrt{4x-8}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)
\(\Leftrightarrow2\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=8\)
\(\Leftrightarrow-3\sqrt{x-2}=8\)(Vô lý)
* Giải phương trình
a. \(\sqrt{45x}-2\sqrt{20x}+2\sqrt{80x}=21\)
b. \(\sqrt{x^2-10x+25}=4\)
* Chứng minh đẳng thức
\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=2\)
Bài 2 :
Ta có : \(\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{5-2\sqrt{5}\sqrt{3}+3}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{15}\right)\left(5+3-2\sqrt{15}\right)\)
\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)\)
\(=2\left(16-15\right)=2.1=2\)
Bài 1 :
a, ĐKXĐ : \(x\ge0\)
Ta có : \(PT\Leftrightarrow3\sqrt{5x}-4\sqrt{5x}+8\sqrt{5x}=21\)
\(\Leftrightarrow7\sqrt{5x}=21\)
\(\Leftrightarrow\sqrt{5x}=3\)
\(\Leftrightarrow x=\dfrac{9}{5}\left(TM\right)\)
Vậy ...
b, Ta có : \(PT\Leftrightarrow\sqrt{\left(x-5\right)^2}=4\)
\(\Leftrightarrow\left|x-5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
Vậy ....
\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\sqrt{4+\sqrt{15}}.\sqrt{4-\sqrt{15}}.\sqrt{4+\sqrt{15}}.\sqrt{2}.\left(\sqrt{5}-\sqrt{3}\right)\)
\(=1.\sqrt{8+2\sqrt{15}}.\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}.\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=2\)
Giải các phương trình sau:
a. \(\sqrt{25x+75}+3\sqrt{x-2}=2\sqrt{x-2}+\sqrt{9x-18}\)
b. \(\sqrt{\left(2x-1\right)^2}=4\)
c. \(\sqrt{\left(2x+1\right)^2}=3x-5\)
d. \(\sqrt{4x-12}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)
a) Ta có: \(\sqrt{25x+75}+3\sqrt{x-2}=2\sqrt{x-2}+\sqrt{9x-18}\)
\(\Leftrightarrow5\sqrt{x+3}+3\sqrt{x-2}=2\sqrt{x-2}+3\sqrt{x-2}\)
\(\Leftrightarrow\sqrt{25x+75}=\sqrt{4x-8}\)
\(\Leftrightarrow25x-4x=-8-75\)
\(\Leftrightarrow21x=-83\)
hay \(x=-\dfrac{83}{21}\)
b) Ta có: \(\sqrt{\left(2x-1\right)^2}=4\)
\(\Leftrightarrow\left|2x-1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
c) Ta có: \(\sqrt{\left(2x+1\right)^2}=3x-5\)
\(\Leftrightarrow\left|2x+1\right|=3x-5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3x-5\left(x\ge-\dfrac{1}{2}\right)\\2x+1=5-3x\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3x=-5-1\\2x+3x=5-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=\dfrac{4}{5}\left(loại\right)\end{matrix}\right.\)
d) Ta có: \(\sqrt{4x-12}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)
\(\Leftrightarrow2\sqrt{x-3}-2\sqrt{x-2}=3\sqrt{x-2}+8\)
\(\Leftrightarrow2\sqrt{x-3}-5\sqrt{x-2}=8\)
\(\Leftrightarrow4\left(x-3\right)+25\left(x-2\right)-20\sqrt{x^2-5x+6}=8\)
\(\Leftrightarrow4x-12+25x-50-8=20\sqrt{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow20\sqrt{\left(x-2\right)\left(x-3\right)}=29x-70\)
\(\Leftrightarrow x^2-5x+6=\dfrac{\left(29x-70\right)^2}{400}\)
\(\Leftrightarrow x^2-5x+6=\dfrac{841}{400}x^2-\dfrac{203}{20}x+\dfrac{49}{4}\)
\(\Leftrightarrow\dfrac{-441}{400}x^2+\dfrac{103}{20}x-\dfrac{25}{4}=0\)
\(\Delta=\left(\dfrac{103}{20}\right)^2-4\cdot\dfrac{-441}{400}\cdot\dfrac{-25}{4}=-\dfrac{26}{25}\)(Vô lý)
vậy: Phương trình vô nghiệm
Bài 1: Tìm x, biết
a)\(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)
b) \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
c)\(\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{4x-4}+\sqrt{x-1}=8\)
d) \(\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}=2\)
a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)
\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
\(\Leftrightarrow4\sqrt{x-3}=20\)
\(\Leftrightarrow x-3=25\)
hay x=28
b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\)
\(\Leftrightarrow x+2=9\)
hay x=7
Bài 1: Giải phương trình:
a, \(\dfrac{3}{4}\sqrt{4x}-\sqrt{4x}+5=\dfrac{1}{4}\sqrt{4x}\)
b,\(\sqrt{3-x}-\sqrt{27-9x}+1,25\sqrt{48-16x}=6\)
Bài 2: Cho biểu thức:
P=\(\left(\dfrac{2}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\dfrac{2}{1-a^2}+1\right)\) (với a\(\ge\)0; a\(\ne\)1)
a, Rút gọn P
b, Tính giá trị của P với a=\(\dfrac{24}{49}\)
c, Tìm a để P=2
Tôi cần gấp hai bài này vào chiều ngày 9 tháng 8 nên mong mọi người giúp đỡ ạ
a) ĐK: \(x\ge0\)
PT \(\Leftrightarrow\sqrt{4x}\left(\dfrac{3}{4}-1-\dfrac{1}{4}\right)+5=0\)
\(\Leftrightarrow2\sqrt{x}.\left(-\dfrac{1}{2}\right)+5=0\)
\(\Leftrightarrow x=25\) (thỏa)
Vậy \(x=25\)
b) Đk: \(x\le3\)
PT \(\Leftrightarrow\sqrt{3-x}-\sqrt{9\left(3-x\right)}+\dfrac{5}{4}\sqrt{16\left(3-x\right)}=6\)
\(\Leftrightarrow\sqrt{3-x}\left(1-\sqrt{9}+\dfrac{5}{4}.\sqrt{16}\right)=6\)
\(\Leftrightarrow\sqrt{3-x}=2\Leftrightarrow x=-1\) (thỏa)
Vậy \(x=-1\)
2:
a:
Sửa đề: \(P=\left(\dfrac{2}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\dfrac{2}{\sqrt{1-a^2}}+1\right)\)
\(P=\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}:\dfrac{2+\sqrt{1-a^2}}{\sqrt{1-a^2}}\)
\(=\dfrac{2+\sqrt{1-a^2}}{\sqrt{1+a}}\cdot\dfrac{\sqrt{1-a^2}}{2+\sqrt{1-a^2}}=\sqrt{\dfrac{1-a^2}{1+a}}\)
\(=\sqrt{1-a}\)
b: Khi a=24/49 thì \(P=\sqrt{1-\dfrac{24}{49}}=\sqrt{\dfrac{25}{49}}=\dfrac{5}{7}\)
c: P=2
=>1-a=4
=>a=-3
1a (đkxđ:\(x\ge0\)) \(\Leftrightarrow\dfrac{-1}{2}.\sqrt{4x}+5=0\) \(\Leftrightarrow\sqrt{4x}=10\) \(\Leftrightarrow x=25\) (t/m)
b (đkxđ:\(x\le3\) ) \(\Leftrightarrow\sqrt{3-x}\left(1-3+1,25.4\right)=6\) \(\Leftrightarrow\sqrt{3-x}=2\) \(\Leftrightarrow x=-1\) (t/m)
\(\sqrt{45x}-2\sqrt{20x}+2\sqrt{80x}=21\)
b.\(\sqrt{x^2-10x+25=4}\)
`a)\sqrt{45x}-2\sqrt{20x}+2\sqrt{80x}=21` `ĐK: x >= 0`
`<=>3\sqrt{5x}-4\sqrt{5x}+8\sqrt{5x}=21`
`<=>7\sqrt{5x}=21`
`<=>\sqrt{5x}=3`
`<=>5x=9<=>x=9/5` (t/m).
`b)\sqrt{x^2-10x+25}=4`
`<=>\sqrt{(x-5)^2}=4`
`<=>|x-5|=4`
`<=>[(x-5=4),(x-5=-4):}`
`<=>[(x=9),(x=1):}`
Giải phương trình:
1. \(2\sqrt{9x-27}-\frac{1}{5}\sqrt{25x-75}-\frac{1}{7}\sqrt{49x-147}=20\)
2.\(\frac{3}{2}\sqrt{5x}+\sqrt{5x}-7=\frac{1}{2}\sqrt{5x}\)
1. ĐKXĐ: \(x\ge3\)
\(2\sqrt{9x-27}-\frac{1}{5}\sqrt{25x-75}-\frac{1}{7}\sqrt{49x-147}=20\)
⇔ \(6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
⇔ \(4\sqrt{x-3}=20\)
⇔ \(\sqrt{x-3}=5\)
⇔ \(x-3=25\)
⇔ \(x=28\left(TMĐKXĐ\right)\)
Vậy....
2. ĐKXĐ: \(x\ge0\)
\(\frac{3}{2}\sqrt{5x}+\sqrt{5x}-7=\frac{1}{2}\sqrt{5x}\)
⇔ \(\frac{3}{2}\sqrt{5x}+\sqrt{5x}-\frac{1}{2}\sqrt{5x}=7\)
⇔ \(2\sqrt{5x}=7\)
⇔ \(\sqrt{5x}=\frac{7}{2}\)
⇔ \(5x=\frac{49}{4}\)
⇔ \(x=\frac{49}{20}\left(TMĐKXĐ\right)\)
Vậy...
giải phương trình:
\(2\sqrt{9x-27}-\frac{1}{5}\sqrt{25x-75}-\frac{1}{7}\sqrt{49x-147}=20\)
đề =
\(2\sqrt{9\left(x-3\right)}-\frac{1}{5}\sqrt{25\left(x-3\right)}-\frac{1}{7}\sqrt{49\left(x-3\right)}=20\)
=>\(6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
=>\(4\sqrt{x-3}=20\)
=>\(\sqrt{x-3}=5\)
=>\(x-3=25\)
=>\(x=28\)