\(24x^2-60x+36-\dfrac{1}{\sqrt{5x-7}}+\dfrac{1}{\sqrt{x-1}}=0\)
Những câu hỏi liên quan
Giải PT
\(24x^2-60x+36-\frac{1}{\sqrt{5x-7}}+\frac{1}{\sqrt{x-1}}=0\)
giải pt
a.\(2\sqrt{x-4}-\dfrac{1}{3}\sqrt{9x-36}=4-\sqrt{x-4}\)
b.\(3\sqrt{x-2}-\sqrt{x^2-4}=0\)
c.\(\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}=-5-x^2+6x\)
a,ĐK: x≥4
Ta có: \(2\sqrt{x-4}-\dfrac{1}{3}\sqrt{9x-36}=4-\sqrt{x-4}\)
\(\Leftrightarrow2\sqrt{x-4}-\sqrt{x-4}=4-\sqrt{x-4}\)
\(\Leftrightarrow2\sqrt{x-4}=4\)
\(\Leftrightarrow\sqrt{x-4}=2\Leftrightarrow x-4=4\Leftrightarrow x=8\left(tm\right)\)
Đúng 1
Bình luận (0)
b, ĐK: x≥2
Ta có: \(3\sqrt{x-2}-\sqrt{x^2-4}=0\)
\(\Leftrightarrow3\sqrt{x-2}-\sqrt{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(3-\sqrt{x+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\3-\sqrt{x+2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\sqrt{x+2}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x+2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=7\end{matrix}\right.\)
Đúng 1
Bình luận (0)
P(dfrac{x-2}{x+2sqrt{x}}+dfrac{1}{sqrt{x+2}}).dfrac{sqrt{x}+1}{sqrt{x}-1} với x 0 và x≠1 A(dfrac{2}{sqrt{x}-2}+dfrac{3}{2sqrt{x+1}}-dfrac{5sqrt{x}-7}{2x-3sqrt{x}-2}):dfrac{2sqrt{x}+3}{5x-10sqrt{x}} với x 0 và x≠4Adfrac{x+1-2sqrt{x}}{sqrt{x-1}}+dfrac{x+sqrt{x}}{sqrt{x}+1} với x≥0 và x≠1V(dfrac{1}{sqrt{x}+2}+dfrac{1}{sqrt{x}-2})dfrac{sqrt{x}+2}{sqrt{x}} với x0 và x≠4A(dfrac{1}{x-1}+dfrac{3sqrt{x}+5}{xsqrt{x}-x-sqrt{x}+1})(dfrac{left(sqrt{x}+1right)^2}{4sqrt{x}}-1)Pdfrac{15sqrt{x}-11}{x...
Đọc tiếp
P=(\(\dfrac{x-2}{x+2\sqrt{x}}\)+\(\dfrac{1}{\sqrt{x+2}}\)).\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) với x >0 và x≠1
A=(\(\dfrac{2}{\sqrt{x}-2}\)+\(\dfrac{3}{2\sqrt{x+1}}\)-\(\dfrac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\)):\(\dfrac{2\sqrt{x}+3}{5x-10\sqrt{x}}\) với x > 0 và x≠4
A=\(\dfrac{x+1-2\sqrt{x}}{\sqrt{x-1}}\)+\(\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\) với x≥0 và x≠1
V=(\(\dfrac{1}{\sqrt{x}+2}\)+\(\dfrac{1}{\sqrt{x}-2}\))\(\dfrac{\sqrt{x}+2}{\sqrt{x}}\) với x>0 và x≠4
A=(\(\dfrac{1}{x-1}\)+\(\dfrac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\))(\(\dfrac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}\)-1)
P=\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}\)+\(\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}\)+\(\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
MỌI NGƯỜI GIÚP ĐỠ MÌNH RÚT GỌN MẤY BIỂU THỨC NÀY VỚI Ạ . EM XIN CẢM ƠN
a: \(P=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: \(=\dfrac{2\left(2\sqrt{x}+1\right)+3\left(\sqrt{x}-2\right)-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}\)
\(=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)
Đúng 0
Bình luận (0)
Giải các phương trình sau:1) sqrt{3x^2+5x+8}-sqrt{3x^2+5x+1}12) x^2-2x-12+4sqrt{left(4-xright)left(2+xright)}03) 3sqrt{x}+dfrac{3}{2sqrt{x}}2x+dfrac{1}{2x}-74) sqrt{x}-dfrac{4}{sqrt{x+2}}+sqrt{x+2}05)left(x-7right)sqrt{dfrac{x+3}{x-7}}x+46) 2sqrt{x-4}+sqrt{x-1}sqrt{2x-3}+sqrt{4x-16}7) sqrt{x+2sqrt{x-1}}+sqrt{x-2sqrt{x-1}}dfrac{x+3}{2}Giúp mình với ajk, mink đang cần gấp
Đọc tiếp
Giải các phương trình sau:
1) \(\sqrt{3x^2+5x+8}-\sqrt{3x^2+5x+1}=1\)
2) \(x^2-2x-12+4\sqrt{\left(4-x\right)\left(2+x\right)}=0\)
3) \(3\sqrt{x}+\dfrac{3}{2\sqrt{x}}=2x+\dfrac{1}{2x}-7\)
4) \(\sqrt{x}-\dfrac{4}{\sqrt{x+2}}+\sqrt{x+2}=0\)
5)\(\left(x-7\right)\sqrt{\dfrac{x+3}{x-7}}=x+4\)
6) \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)
7) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\)
Giúp mình với ajk, mink đang cần gấp
25 tháng 7 2023 lúc 12:38
2) Tính: (Giải chi tiết từng bước)
a) \(2\sqrt{125}+\dfrac{3}{2}\sqrt{80}-\sqrt{180}-\dfrac{2}{7}\sqrt{245}\)
b) \(\sqrt{11-4\sqrt{7}}-\sqrt{16+6\sqrt{7}}\)
3) Tìm x, biết:
a) \(\sqrt{\left(x-1\right)^2}=4\)
b) \(\sqrt{36x^2-60x+25}=4\)
Bài 2:
a) \(2\sqrt{125}+\dfrac{3}{2}\sqrt{80}-\sqrt{180}-\dfrac{2}{7}\sqrt{245}\)
\(=2\sqrt{5^2\cdot5}+\dfrac{3}{2}\sqrt{4^2\cdot5}-\sqrt{6^2\cdot5}-\dfrac{2}{7}\sqrt{7^2\cdot5}\)
\(=10\sqrt{5}+\dfrac{3\cdot4}{2}\sqrt{5}-6\sqrt{5}-\dfrac{2\cdot7}{7}\sqrt{5}\)
\(=10\sqrt{5}+6\sqrt{6}-6\sqrt{5}-2\sqrt{5}\)
\(=8\sqrt{5}\)
b) \(\sqrt{11-4\sqrt{7}}-\sqrt{16+6\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}\right)^2-2\cdot2\cdot\sqrt{7}+2^2}-\sqrt{\left(\sqrt{7}\right)^2+2\cdot3\cdot\sqrt{7}+3^2}\)
\(=\sqrt{\left(\sqrt{7}-2\right)^2}-\sqrt{\left(\sqrt{7}+3\right)^2}\)
\(=\sqrt{7}-2-\sqrt{7}-3\)
\(=-5\)
Đúng 2
Bình luận (0)
\(2a,\\ 2\sqrt{125}+\dfrac{3}{2}.\sqrt{80}-\sqrt{180}-\dfrac{2}{7}\sqrt{245}\\ =2\sqrt{5^2.5}+\dfrac{3}{2}.\sqrt{4^2.5}-\sqrt{6^2.5}-\dfrac{2}{7}.\sqrt{7^2.5}\\ =2.5.\sqrt{5}+\dfrac{3}{2}.4.\sqrt{5}-6\sqrt{5}-\dfrac{2}{7}.7\sqrt{5}\\ =10\sqrt{5}+6\sqrt{5}-6\sqrt{5}-2\sqrt{5}=8\sqrt{5}\)
Đúng 0
Bình luận (0)
3:
a: =>|x-1|=4
=>x-1=4 hoặc x-1=-4
=>x=-3 hoặc x=5
b: =>|6x-5|=4
=>6x-5=4 hoặc 6x-5=-4
=>6x=1 hoặc 6x=9
=>x=1/6 hoặc x=3/2
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
1) a,\(\sqrt{x}-1=5\Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\) b, \(\dfrac{1}{2}\sqrt{5x}< 1\Leftrightarrow\sqrt{5x}< 2\Leftrightarrow5x=4\Leftrightarrow x=\dfrac{4}{5}\)
2) A=bấm máy tính B=\(\left(\sqrt{28}-\sqrt{7}+\sqrt{12}\right):\sqrt{7}\Leftrightarrow4-1+\sqrt{7\cdot12}\Leftrightarrow3+\sqrt{84}\)
Giải phương trình sau:
a) \(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
b) \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
c) \(2x-x^2+\sqrt{6x^2-12x+7}=0\)
d) \(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6\)
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
Đúng 1
Bình luận (0)
Tìm x
1) \(\sqrt{\dfrac{3x-1}{x+2}}=2\)
2)\(\sqrt{\dfrac{5x-7}{2x- 1}}=2\)
3)\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
4) \(\dfrac{\sqrt{x}-3}{\sqrt{x}+2}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
1: \(\Leftrightarrow\dfrac{3x-1}{x+2}=4\)
=>4x+8=3x-1
=>x=-9
2: \(\Leftrightarrow\dfrac{5x-7}{2x-1}=4\)
=>8x-4=5x-7
=>3x=-3
=>x=-1
3: ĐKXD: x>=0
\(PT\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
=>\(x+\sqrt{x}-6=x-1\)
=>căn x=-1+6=5
=>x=25
4: ĐKXĐ: x>=0
PT =>\(\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
=>x-2*căn x-3=x-4
=>-2căn x-3=-4
=>2căn x+3=4
=>2căn x=1
=>căn x=1/2
=>x=1/4
Đúng 0
Bình luận (0)
26 tháng 10 2021 lúc 9:41
6) \(\sqrt{x^2+12x+36}=-x-6\)
7) \(\sqrt{9x^2-12x+4}=3x-2\)
8) \(\sqrt{16-24x+9x^2}=2x-10\)
9) \(\sqrt{x^2-6x+9}==2x-3\)
10) \(\sqrt{x^2-3x+\dfrac{9}{4}}=\dfrac{3}{x}x-4\)
6) ĐKXĐ: \(x\le-6\)
\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)
\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)
Vậy \(x\le-6\)
7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)
\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)
\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)
Vậy \(x\ge\dfrac{2}{3}\)
8) ĐKXĐ: \(x\ge5\)
\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)
\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)
9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
Đúng 1
Bình luận (0)