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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
Bài 2:
1: =>5x+1=6/7 hoặc 5x+1=-6/7
=>5x=-1/7 hoặc 5x=-13/7
=>x=-1/35 hoặc x=-13/35
2: =>x-1=4
=>x=5
3: =>3x-1=3
=>3x=4
=>x=4/3
4: \(\Leftrightarrow\dfrac{5}{x+3}=\dfrac{-5}{6}+\dfrac{1}{2}=\dfrac{-5+3}{6}=\dfrac{-2}{6}=\dfrac{-1}{3}\)
=>x+3=-15
=>x=-18
7: \(\Leftrightarrow2^{2x+1}+2^{2x+6}=264\)
=>2^2x+1*(1+2^5)=264
=>2^2x+1=8
=>2x+1=3
=>x=1
9: =>x^4=8x
=>x^4-8x=0
=>x=2
Bài 1
a, \(D=1-\left|2x-3\right|\)
Ta có : \(\left|2x-3\right|\ge0\)
\(\Rightarrow1-\left|2x-3\right|\le1\)
Dấu "=" xảy ra khi \(\left|2x-3\right|=0\)
\(\Leftrightarrow2x-3=0\)
\(\Leftrightarrow2x=3\)
\(\Leftrightarrow x=3:2=\dfrac{3}{2}\)
\(b,\) Ta có : \(\left|10-5x\right|\ge0\Rightarrow\left|10-5x\right|+14,2\ge14,3\Rightarrow-\left|10-5x\right|-14,2\le-14,2\)
Dấu "=" xảy ra khi \(-\left|10-5x\right|=0\)
\(\Leftrightarrow10-5x=0\)
\(\Leftrightarrow5x=10\)
\(\Leftrightarrow x=10:5=2\)
Vậy \(Emax=-14,2\Leftrightarrow x=2\)
\(c,\) Ta có : \(\left|5x-2\right|\ge0\)
\(\left|3y-12\right|\ge0\)
⇒ \(\left|5x-2\right|+\left|3y+12\right|-4\ge-4\)
⇒ \(4-\left|5x-2\right|-\left|3y+12\right|\le4\)
Dấu "=" xảy ra khi \(\left[{}\begin{matrix}\left|5x-2\right|=0\\\left|3y+12\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=2\\3y=-12\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\y=-4\end{matrix}\right.\)
\(d,\) \(A=5-3\left(2x-1\right)^2\)
Ta có : \(\left(2x-1\right)^2\ge0\)
\(\Rightarrow3.\left(2x-1\right)^2\ge0\)
\(\Rightarrow3.\left(2x-1\right)^2-5\ge-5\)
\(\Rightarrow5-3\left(2x-1\right)^2\le5\)
Dấu "=" xảy ra khi \(\left(2x-1\right)^2=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(Amax=5\Leftrightarrow x=\dfrac{1}{2}\)
a) \(0,75:4,5=\dfrac{1}{15}:\left(2x\right)\)
\(\Rightarrow\) \(\dfrac{1}{6}=\dfrac{1}{30}:x\)
\(\Rightarrow\) \(x=\dfrac{1}{5}\)
a. \(0,75:4,5=\dfrac{1}{15}:\left(2x\right)\)
\(\Leftrightarrow\dfrac{1}{15}:\left(2x\right)=0,75:4,5\)
\(\Rightarrow\dfrac{1}{15}:\left(2x\right)=\dfrac{1}{6}\)
\(\Rightarrow2x=\dfrac{1}{15}:\dfrac{1}{6}=\dfrac{2}{5}\)
\(\Rightarrow x=\dfrac{2}{5}:2=\dfrac{1}{5}\)
Vậy...
b. \(\dfrac{-5}{x-2}=\dfrac{3}{-9}\)
\(\Leftrightarrow\left(x-2\right).3=\left(-5\right).\left(-9\right)\)
\(\Rightarrow\left(x-2\right).3=45\)
\(\Rightarrow\left(x-2\right)=45:3=15\)
\(\Rightarrow x=15+2=17\)
Vậy...
c. \(\dfrac{-2}{3}:x=\dfrac{1}{2}:\dfrac{3}{4}\)
\(\Rightarrow\dfrac{-2}{3}:x=\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{-2}{3}:\dfrac{2}{3}=-1\)
Vậy...
Bài 7:
x/1=z/2 nên x/6=z/12
=>x/6=y/9=z/12
=>x/2=y/3=z/4
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{27}{9}=3\)
=>x=6; y=9; z=12
\(\dfrac{x+2}{0,5}=\dfrac{2x+1}{2}\)
\(\Leftrightarrow\left(x+2\right).2=\left(2x+1\right).0,5\)
\(\Leftrightarrow2x+4=x+0,5\)
\(\Leftrightarrow x=-3,5\)
Vậy...
\(\dfrac{x+2}{0,5}=\dfrac{2x+1}{2}\)
\(\Leftrightarrow\dfrac{4.\left(x+2\right)}{2}=\dfrac{2x+1}{2}\)
\(\Rightarrow4x+8=2x+1\)
\(\Leftrightarrow4x-2x=1-8\)
\(\Leftrightarrow2x=-7\)
\(\Leftrightarrow x=\dfrac{-7}{2}\)
Vậy \(x=\dfrac{-7}{2}\)
Đặt \(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{t}=k\)
=>\(x=yk;y=kz;z=kt\)
Ta có: \(\left(\dfrac{x+y+z}{y+z+t}\right)^3\)
\(=\left(\dfrac{yk+kz+kt}{y+z+t}\right)^3=\left(\dfrac{k\left(y+z+t\right)}{y+z+t}\right)^3=k^3\left(1\right)\)
Ta có: \(\dfrac{x}{t}=\dfrac{yk}{t}=\dfrac{k^2z}{t}=\dfrac{k^3t}{t}=k^3\left(2\right)\)
Từ (1) và (2) suy ra \(\left(\dfrac{x+y+z}{y+z+t}\right)^3=\dfrac{x}{t}\)
Vậy \(\left(\dfrac{x+y+z}{y+z+t}\right)^3=\dfrac{x}{t}\)
Cho mk 1 like nhé ^_^
\(\dfrac{x+2}{x-4}=\dfrac{2x+3}{x-6}\) (ĐKXĐ: \(x\ne4;x\ne6\))
\(\Rightarrow\left(x+2\right)\left(x-6\right)=\left(2x+3\right)\left(x-4\right)\)
\(\Rightarrow x^2-6x+2x-12=2x^2-8x+3x-12\)
\(\Rightarrow x^2-x=0\)
\(\Rightarrow x\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)