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Bài 2 :
a, \(A=\left|2x-4\right|+2\ge2\)
Dấu ''='' xảy ra khi x = 2
Vậy GTNN A là 2 khi x = 2
b, \(B=\left|x+2\right|-3\ge-3\)
Dấu ''='' xảy ra khi x = -2
Vậy GTNN B là -3 khi x = -2
Hiền Hòa mac xac ta do ba tam nhieu chuyen lelelel
a) -4/5 + 5/2x = -3/10
5/2x = -3/10 + 4/5
5/2x = 1/5
5/2x = 1/2
x = 1/2 : 5/2
x = 1/5
b) 4/3 + 5/8 : x = 1/12
5/8x = 1/12 - 4/3
5/8x = -5/4
5 = -5/4.8x
5 = -10x
5/-10 = x
-1/2 = x
x = -1/2
c) (x - 1/3)(x - 2/5) = 0
x - 1/3 = 0 hoặc x - 2/5 = 0
x = 0 + 1/3 x = 0 + 2/5
x = 1/3 x = 2/5
\(a,\dfrac{12}{5}=\dfrac{x}{1,5}\Rightarrow x=\dfrac{12\cdot1,5}{5}=3,6\\ b,\dfrac{x}{5}=\dfrac{3}{20}\Rightarrow x=\dfrac{5\cdot3}{20}=\dfrac{3}{4}\\ c,\dfrac{4}{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{4\cdot9}{10}=\dfrac{18}{5}\\ d,\Rightarrow\dfrac{x}{15}=\dfrac{60}{x}\Rightarrow x^2=60\cdot15=900\Rightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\\ 2,\)
a, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x+y-z}{3+5-6}=\dfrac{8}{2}=4\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=20\\z=24\end{matrix}\right.\)
b, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x-y+z}{3-5+6}=\dfrac{-4}{4}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-6\end{matrix}\right.\)
c, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{2y}{10}=\dfrac{3z}{18}=\dfrac{x-2y+3z}{3-10+18}=\dfrac{-33}{11}=-3\\ \Rightarrow\left\{{}\begin{matrix}x=-9\\y=-15\\z=-18\end{matrix}\right.\)
d, Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=k\Rightarrow x=3k;y=5k;z=6k\)
\(x^2-4y^2+2z^2=-475\\ \Rightarrow9k^2-100k^2+72z^2=-475\\ \Rightarrow-19k^2=-475\\ \Rightarrow k^2=25\Rightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=15;y=25;z=30\\x=-15;y=-25;z=-30\end{matrix}\right.\)
a, \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}\Rightarrow}\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
b. \(\left(x^2+1\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x^2=-1\left(Voly\right)\\x=4\end{cases}\Rightarrow x=4}\)
c, \(2x^2-\frac{1}{3}x=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)
d, \(\left(\frac{4}{5}\right)^{5x}=\left(\frac{4}{5}\right)^7\)
\(\Rightarrow5x=7\)
\(\Rightarrow x=\frac{7}{5}\)
e, Ta có: \(A=\frac{x+5}{x-2}=\frac{\left(x-2\right)+7}{x-2}=1+\frac{7}{x-2}\)
Để A ∈ Z <=> (x - 2) ∈ Ư(7) = { ±1; ±7 }
| x - 2 | 1 | -1 | 7 | -7 |
| x | 3 | 1 | 9 | -5 |
Vậy....
a) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)
Vậy : ....
b) \(\left(x^2+1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=-1\left(loại\right)\\x=4\end{cases}}\)
c) \(2x^2-\frac{1}{3}x=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)
Vậy :...
Bài 1 :
\(\frac{x-1}{x-5}=\frac{6}{7}\Leftrightarrow7x-7=6x-30\)
\(\Leftrightarrow x=-23\)
\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)ĐK : \(x\ne1;-7\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-1\right)\)
\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)
\(\Leftrightarrow2x-10=0\Leftrightarrow x=5\)
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
