Tìm x biết:
a) \(\dfrac{14}{5}.x-50=34\)
b) \(\left(25\%-2\dfrac{1}{2}\right).2x-\dfrac{5}{4}.2^4=\dfrac{525}{100}\)
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
\(\text{Tìm x, biết:}\)
\(a\)) \(20\text{%}x-x+\dfrac{1}{5}=\dfrac{3}{4}\)
\(b\)) \(\dfrac{2x+1}{3}=\dfrac{x-5}{2}\)
\(c\)) \(\left(x-\dfrac{3}{4}\right)\left(4+3x\right)=0\)
\(d\)) \(x-\dfrac{1}{3}x+\dfrac{1}{5}x=\dfrac{-26}{5}\)
\(e\)) \(50\text{%}x+\dfrac{2}{3}x=x-5\)
\(g\)) \(\dfrac{2}{3}\left(x+\dfrac{9}{5}\right)-\dfrac{3}{10}.\left(5x-\dfrac{1}{3}\right)=\dfrac{7}{15}\)
câu c) mang tính mua vui hay gì hả bn
mếu thật thì x=0,x=số nào cx đc(câu trả lời này mang tính mua vui thôi nhé)
1. \(\dfrac{5\left(x-1\right)+2}{6}-\dfrac{7x-1}{4}=\dfrac{2\left(2x+1\right)}{7}-5\)
2. \(x-\dfrac{3\left(x+30\right)}{15}-24\dfrac{1}{2}=\dfrac{7x}{10}-\dfrac{2\left(10x+2\right)}{5}\)
3. \(14\dfrac{1}{2}-\dfrac{2\left(x+3\right)}{5}=\dfrac{3x}{2}-\dfrac{2\left(x-7\right)}{3}\)
4. \(\dfrac{x+1}{3}+\dfrac{3\left(2x+1\right)}{4}=\dfrac{2x+3\left(x+1\right)}{6}+\dfrac{7+12x}{12}\)
5. \(\dfrac{3\left(2x-1\right)}{4}-\dfrac{3x+1}{10}+1=\dfrac{2\left(3x+2\right)}{5}\)
6. \(x-\dfrac{3}{17}\left(2x-1\right)=\dfrac{7}{34}\left(1-2x\right)+\dfrac{10x-3}{2}\)
7. \(\dfrac{3\left(x-3\right)}{4}+\dfrac{4x-10,5}{10}=\dfrac{3\left(x+1\right)}{5}+6\)
8. \(\dfrac{2\left(3x+1\right)+1}{4}-5=\dfrac{2\left(3x-1\right)}{5}-\dfrac{3x+2}{10}\)
o) \(\dfrac{\left(-1\right)^6.3^5.4^3}{9^2.2^5}\)
s) \(\dfrac{\dfrac{2}{7}+\dfrac{2}{5}+\dfrac{2}{17}-\dfrac{2}{25}}{\dfrac{3}{14}+\dfrac{3}{10}+\dfrac{3}{34}-\dfrac{3}{50}}\)
t) \(\sqrt{\dfrac{4}{9}}\) - \(\dfrac{1}{2}\): \(\left|\dfrac{-2}{3}\right|\)
Mg giải gấp giúp mình với
o: \(\dfrac{\left(-1\right)^6\cdot3^5\cdot4^3}{9^2\cdot2^5}=\dfrac{3^5\cdot2^6}{2^5\cdot3^4}=\dfrac{3^5}{3^4}\cdot\dfrac{2^6}{2^5}=3\cdot2=6\)
s: \(\dfrac{\dfrac{2}{7}+\dfrac{2}{5}+\dfrac{2}{17}-\dfrac{2}{25}}{\dfrac{3}{14}+\dfrac{3}{10}+\dfrac{3}{34}-\dfrac{3}{50}}\)
\(=\dfrac{2\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{25}\right)}{\dfrac{3}{2}\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{25}\right)}\)
\(=2:\dfrac{3}{2}=\dfrac{4}{3}\)
t: \(\sqrt{\dfrac{4}{9}}-\dfrac{1}{2}:\left|-\dfrac{2}{3}\right|\)
\(=\dfrac{2}{3}-\dfrac{1}{2}:\dfrac{2}{3}\)
\(=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8-9}{12}=-\dfrac{1}{12}\)
Phân tích đa thức \(18x^3-\dfrac{8}{25}x\) thành nhân tử
a. \(\dfrac{2}{25}x\left(9x^2-4\right)=\dfrac{2}{25}x\left(3x-2\right)\left(3x+2\right)\)
b. \(2x\left(9x^2-\dfrac{4}{25}\right)=2x\left(3x-\dfrac{2}{5}\right)\left(3x+\dfrac{2}{5}\right)\)
Cách phân tích nào đúng, a hay b. Giải thích vì sao?
Giải các phương trình sau :
1.\(\dfrac{14}{3x-12}-\dfrac{2+x}{x-4}=\dfrac{3}{8-2x}-\dfrac{5}{6}\)
2.\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\)
3.\(\dfrac{x+5}{x^2-5x}-\dfrac{x+25}{2x^2-50}=\dfrac{x-5}{2x^2+10x}\)
4.\(\dfrac{6x_{ }+1}{x^2-7x+10}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)
5.\(\dfrac{2}{x^2-4}-\dfrac{x-1}{x\left(x-2\right)}+\dfrac{x-4}{x\left(x+2\right)}=0\)
6.\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\)
Câu 2:
ĐKXĐ: \(\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Rightarrow \left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(1\right)\)
\(\left(1\right):\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)
\(\Leftrightarrow 12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)
\(\Leftrightarrow 12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)
\(\Leftrightarrow12x+12=0\\ \Leftrightarrow12x=-12\\ \Leftrightarrow x=-1\left(TM\right)\)
Vậy \(S=\left\{-1\right\}\)
Tìm x liên quan đến lũy thừa:
1, \(\left(3x-\dfrac{1}{5}\right)^2=\left(\dfrac{-3}{25}\right)^2\)
2, \(\left(2x-\dfrac{1}{3}\right)^2=\left(\dfrac{-2}{9}\right)^2\)
3, \(\left(\dfrac{1}{3}-x\right)^2=\dfrac{9}{25}\)
4, \(\left(5-x\right)^2=25\)
1: \(\left(3x-\dfrac{1}{5}\right)^2=\left(-\dfrac{3}{25}\right)^2\)
=>3x-1/5=3/25 hoặc 3x-1/5=-3/25
=>3x=8/25 hoặc 3x=2/25
=>x=8/75 hoặc x=2/75
2: \(\left(2x-\dfrac{1}{3}\right)^2=\left(-\dfrac{2}{9}\right)^2\)
=>2x-1/3=2/9 hoặc 2x-1/3=-2/9
=>2x=5/9 hoặc 2x=1/9
=>x=5/18 hoặc x=1/18
1) giải pt :
a) \(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)
b) \(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=-5\)
2) giải pt :
a) \(\left(5x+1\right)^2=\left(3x-2\right)^2\)
b) \(\left(x+2\right)^3=\left(2x+1\right)^3\)
c) \(\left(x+3\right)^4+\left(x+5\right)^4=2\)
d) \(x^4-3x^3+4x^2-3x+1=0\)
1)
\(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)
\(\Leftrightarrow\dfrac{x-5}{100}+1+\dfrac{x-4}{101}+1+\dfrac{x-3}{102}+1=\dfrac{x-100}{5}+1+\dfrac{x-101}{4}+1+\dfrac{x-102}{3}+1\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}=\dfrac{x-105}{5}+\dfrac{x-105}{4}+\dfrac{x-105}{3}+\dfrac{x-105}{2}\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}-\dfrac{x-105}{5}-\dfrac{x-105}{4}-\dfrac{x-105}{3}-\dfrac{x-105}{2}=0\)
\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)\(\Leftrightarrow105-x=0\)
\(\Leftrightarrow x=105\)
b)
\(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=0\)
\(\Leftrightarrow\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)
\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{20-x}{27}+\dfrac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\right)=0\)
\(\Leftrightarrow50-x=0\)
\(\Leftrightarrow x=50\)
2)
\(\left(5x+1\right)^2=\left(3x-2\right)^2\)
\(\Leftrightarrow\left|5x+1\right|=\left|3x-2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=3x-2\\5x+1=-3x+2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{8}\end{matrix}\right.\)
b) \(\left(x+2\right)^3=\left(2x+1\right)^3\)
\(\Leftrightarrow x^3+6x^2+12x+8=8x^3+12x^2+6x+1\)
\(\Leftrightarrow-7x^3-6x^2+6x+7=0\)
\(\Leftrightarrow-7x^3+7x^2-13x^2+13x-7x+7=0\)
\(\Leftrightarrow-7x^2\left(x-1\right)-13x\left(x-1\right)-7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-7x^2-13x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-7x^2-13x-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x^2+\dfrac{13}{7}x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x+\dfrac{13}{14}\right)^2-\dfrac{169}{196}=0\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow x=1\)
c. \(\left(x+3\right)^4+\left(x+5\right)^4=2\)
Đặt: \(y=x+4\), ta có:
\(\left(y-1\right)^4+\left(y+1\right)^4=2\)
\(\Leftrightarrow y^4-4y^3+6y^2-4y+1+y^4+4y^3+6y^2+4y+1=2\)
\(\Leftrightarrow2y^4+12y^2=0\)
\(\Leftrightarrow2y^2\left(y^2+6\right)=0\)
\(\Leftrightarrow y=0\)
\(\Leftrightarrow x=-4\)
d) \(x^4-3x^3+4x^2-3x+1=0\)
\(\Leftrightarrow x^4-x^3-2x^3+2x^2+2x^2-2x-x+1=0\)
\(\Leftrightarrow x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-2x^2+2x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-x^2-x^2+x+x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x-1\right)-x\left(x-1\right)+\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^2-x+1\right)=0\)
\(\Leftrightarrow x=1\)
Tìm x biết:
\(a,3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\)
\(b,\dfrac{1}{3}+\dfrac{2}{3}:x=-7\)
\(c,\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(d,\left(2x-3\right)\left(6-2x\right)=0\)
\(e,x:\dfrac{3}{4}+\dfrac{1}{4}=-\dfrac{2}{3}\)
\(f,\dfrac{-2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}\)
\(g,2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\)
\(h,\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)
\(i,\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}-\left(-1\right)=\dfrac{1}{3}\)
\(j,\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(k,\dfrac{1}{4}+\dfrac{1}{3}:\left(2x-1\right)=-5\)
\(l,\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)
\(m,3\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\)
\(n,60\%x+\dfrac{2}{3}x=\dfrac{1}{3}.6\dfrac{1}{3}\)
\(p,-5\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(q,3\left(x-\dfrac{1}{2}\right)-5\left(x+\dfrac{3}{5}\right)=-x+\dfrac{1}{5}\)
a: =>1/2x=7/2-2/3=21/6-4/6=17/6
=>x=17/3
b: =>2/3:x=-7-1/3=-22/3
=>x=2/3:(-22/3)=-1/11
c: =>1/3x+2/5x-2/5=0
=>11/15x=2/5
hay x=6/11
d: =>2x-3=0 hoặc 6-2x=0
=>x=3/2 hoặc x=3