Tìm x: \(\left(x-\dfrac{1}{5}\right):\left(x-1\dfrac{6}{7}\right)< 0\)
HELP ME!
Tìm x :
1) \(\left(-0,75x+\dfrac{5}{2}\right).\dfrac{4}{7}-\left(-\dfrac{1}{3}\right)=-\dfrac{5}{6}\)
2) \(\left(4x-9\right)\left(2,5+\dfrac{-7}{3}x\right)=0\)
3) \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
4)\(\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=\dfrac{-64}{125}\)
3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
Tìm x:
\(\dfrac{\sqrt{x}+3}{\sqrt{x}-1}\left(\dfrac{-2x+6}{\sqrt{x}-1}\right)=0\)
Help me plsss
ĐKXĐ: x>=0; x<>1
PT =>\(\dfrac{\left(\sqrt{x}+3\right)\left(-2x+6\right)}{\left(\sqrt{x}-1\right)^2}=0\)
=>6-2x=0
=>x=3
Tìm x: \(\left(x-\dfrac{1}{5}\right):\left(x-1\dfrac{6}{7}\right)< 0\)
\(\dfrac{\left(x-\dfrac{1}{5}\right)}{x-1\dfrac{6}{7}}< 0\)
=>\(\dfrac{x-\dfrac{1}{5}}{x-\dfrac{13}{7}}< 0\)
TH1: x-1/5>0 và x-13/7<0
=>x>1/5 và x<13/7
=>1/5<x<13/7
TH2: x-1/5<0 và x-13/7>0
=>x>13/7 hoặc x<1/5
=>Loại
\(\left(x-\dfrac{1}{5}\right):\left(x-1\dfrac{6}{7}\right)< 0\)
\(\Rightarrow\left(x-\dfrac{1}{5}\right):\left(x-\dfrac{13}{7}\right)< 0\)
\(\Rightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{1}{5}< 0\\x-\dfrac{13}{7}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{1}{5}>0\\x-\dfrac{13}{7}< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{1}{5}\\x>\dfrac{13}{7}\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{1}{5}\\x< \dfrac{13}{7}\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{1}{5}\\x>\dfrac{13}{7}\end{matrix}\right.\\x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow\dfrac{13}{7}< x< \dfrac{1}{5}\)
Để giải bất phương trình này, ta cần tìm khoảng giá trị của x thỏa mãn bất phương trình đã cho.Đầu tiên, ta cần tìm điểm mà tử số và mẫu số của biểu thức đạt giá trị 0.Tử số đạt giá trị 0 khi x - 15 = 0, tức x = 15.Mẫu số đạt giá trị 0 khi x - 167 = 0, tức x = 167.Tiếp theo, ta cần xác định khoảng giá trị nằm giữa hai điểm đã tìm được. Ta chọn một điểm x bất kỳ trong khoảng giữa 15 và 167, ví dụ x = 100.Đặt x = 100 vào biểu thức đã cho:(100 - 15) : (100 - 167) < 085 : (-67) < 0-85/67 < 0Vì biểu thức đạt giá trị âm, nên ta có: (x - 15) : (x - 167) < 0
Tìm x biết:
\(a,\left(x-\dfrac{3}{4}\right)+50\%=\dfrac{1}{6}\)
\(b,\dfrac{1}{2}x-\dfrac{5}{6}x=\dfrac{7}{2}\)
\(c,\left(4-x\right)\left(3x+5\right)=0\)
\(d,\dfrac{x}{16}=\dfrac{50}{32}\)
\(e,\left(2x-3\right)+\dfrac{3}{2}=-\dfrac{1}{4}\)
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
Tính nhanh
A=\(\dfrac{1}{\left(x+1\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+9\right)}+\dfrac{1}{\left(x+9\right)\left(x+11\right)}\)
help me please
\(A=\dfrac{1}{\left(x+1\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+9\right)}+\dfrac{1}{\left(x+9\right)\left(x+11\right)}\)\(A=\dfrac{1}{2}\left(\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+7}+\dfrac{1}{x+7}-\dfrac{1}{x+9}+\dfrac{1}{x+9}-\dfrac{1}{x+11}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{1}{x+1}-\dfrac{1}{x+11}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{x+11}{\left(x+1\right)\left(x+11\right)}-\dfrac{x+1}{\left(x+1\right)\left(x+11\right)}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{x+11-x-1}{\left(x+1\right)\left(x+11\right)}\right)=\dfrac{1}{2}.\dfrac{10}{\left(x+1\right)\left(x+11\right)}=\dfrac{10}{2\left(x+1\right)\left(x+11\right)}\)
Tìm x :
a) \(\dfrac{2x-3}{3}+\dfrac{-3}{2}=\dfrac{5-3x}{6}-\dfrac{1}{3}\)
b) \(\dfrac{2}{3x}-\dfrac{3}{12}=\dfrac{4}{5}-\left(\dfrac{7}{x}-2\right)\)
c) \(\dfrac{x+2014}{2}+\dfrac{2x+4028}{7}=\dfrac{x+2009}{5}+\dfrac{x+2020}{6}\)
d)\(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+18\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
Help me , bn nào giải đc bài nò thì giải nha !!! =))
a: =>4x-6-9=5-3x-3
=>4x-15=-3x+2
=>7x=17
hay x=17/7
b: \(\Leftrightarrow\dfrac{2}{3x}-\dfrac{1}{4}=\dfrac{4}{5}-\dfrac{7}{x}+2\)
=>2/3x+21/3x=4/5+2+1/4=61/20
=>23/3x=61/20
=>3x=23:61/20=460/61
hay x=460/183
Rút gọn biểu thức sau
A=\(\dfrac{1}{x-1}\sqrt{75\left(x-1\right)^3}\left(x>1\right)
\)
B=\(5\sqrt{4x}-3\sqrt{\dfrac{100x}{9}}-\dfrac{4}{x}\sqrt{\dfrac{x^3}{4}}\left(x>0\right)
\)
C=\(x-4+\sqrt{16-8x+x^2}\left(x>4\right)\)
Help me
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
I : Giải các phường trình sau
a) \(\left(3x-2\right)\left(\dfrac{2\left(x+3\right)}{7}-\dfrac{4x-3}{5}\right)=0\)
b) \(\left(x-\dfrac{3}{4}\right)^2+\left(x-\dfrac{3}{4}\right)\left(x-\dfrac{1}{2}\right)=0\)
c) \(\dfrac{12}{9-x^2}+\dfrac{2}{x-3}+\dfrac{3}{x+3}=1\)
d) \(\dfrac{1}{x-1}+\dfrac{2}{x^2+x+1}=\dfrac{3x^2}{x^3-1}\)
help me
Tìm x, biết:
a) \(\dfrac{-3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}\left(x+1\right)\)
b) \(\left(6-3\sqrt{x}\right)\left(\left|x\right|-7\right)=0\)
a: \(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}\left(x+1\right)\)
=>\(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}x+\dfrac{1}{2}\)
=>\(-\dfrac{3}{2}x-\dfrac{1}{2}x=\dfrac{1}{2}-\dfrac{1}{4}\)
=>\(-2x=\dfrac{1}{4}\)
=>\(2x=-\dfrac{1}{4}\)
=>\(x=-\dfrac{1}{4}:2=-\dfrac{1}{8}\)
b: ĐKXĐ: x>=0
\(\left(6-3\sqrt{x}\right)\left(\left|x\right|-7\right)=0\)
=>\(\left\{{}\begin{matrix}6-3\sqrt{x}=0\\\left|x\right|-7=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3\sqrt{x}=6\\\left|x\right|=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=2\\\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-7\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)