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1)
vì | 1 - 2x | \(\ge\)0 \(\Rightarrow\)| 1 - 2x | - 2009 \(\ge\)-2009
\(\Rightarrow\)GTNN của A là -2009 khi | 1 - 2x | = 0 hay x = \(\frac{1}{2}\)
2)
\(\frac{x}{y}=\frac{2}{5}\Rightarrow\frac{x}{2}=\frac{y}{5}=\frac{x+y}{2+5}=\frac{-21}{7}=-3\)
\(\Rightarrow x=\left(-3\right).2=-6;y=\left(-3\right).5=-15\)
3)
2225 = ( 23 )75 = 875
3150 = ( 32 )75 = 975
vì 875 < 975 nên 2225 < 3150
a)
\((3x-7)^5=0\Rightarrow 3x-7=0\Rightarrow x=\frac{7}{3}\)
b)
\(\frac{1}{4}-(2x-1)^2=0\)
\(\Leftrightarrow (2x-1)^2=\frac{1}{4}=(\frac{1}{2})^2=(-\frac{1}{2})^2\)
\(\Rightarrow \left[\begin{matrix} 2x-1=\frac{1}{2}\\ 2x-1=\frac{-1}{2}\end{matrix}\right.\Rightarrow \Rightarrow \left[\begin{matrix} x=\frac{3}{4}\\ x=\frac{1}{4}\end{matrix}\right.\)
c)
\(\frac{1}{16}-(5-x)^3=\frac{31}{64}\)
\(\Leftrightarrow (5-x)^3=\frac{1}{16}-\frac{31}{64}=\frac{-27}{64}=(\frac{-3}{4})^3\)
\(\Leftrightarrow 5-x=\frac{-3}{4}\)
\(\Leftrightarrow x=\frac{23}{4}\)
d)
\(2x=(3,8)^3:(-3,8)^2=(3,8)^3:(3,8)^2=3,8\)
\(\Rightarrow x=3,8:2=1,9\)
e)
\((\frac{27}{64})^9.x=(\frac{-3}{4})^{32}\)
\(\Leftrightarrow [(\frac{3}{4})^3]^9.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow (\frac{3}{4})^{27}.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow x=(\frac{3}{4})^{32}:(\frac{3}{4})^{27}=(\frac{3}{4})^5\)
f)
\(5^{(x+5)(x^2-4)}=1\)
\(\Leftrightarrow (x+5)(x^2-4)=0\)
\(\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2=4=2^2=(-2)^2\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=-5\\ x=\pm 2\end{matrix}\right.\)
g)
\((x-2,5)^2=\frac{4}{9}=(\frac{2}{3})^2=(\frac{-2}{3})^2\)
\(\Rightarrow \left[\begin{matrix} x-2,5=\frac{2}{3}\\ x-2,5=\frac{-2}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{19}{6}\\ x=\frac{11}{6}\end{matrix}\right.\)
h)
\((2x+\frac{1}{3})^3=\frac{8}{27}=(\frac{2}{3})^3\)
\(\Rightarrow 2x+\frac{1}{3}=\frac{2}{3}\Rightarrow x=\frac{1}{6}\)
a)x-3/x+5=5/7 suy ra 7.(x-3) = 5(x+5)
Tương đương : 7x - 21 = 5x + 25
7x - 5x = 25 + 21 = 46
2x = 46 suy ra : x = 46/2 = 23
Vậy x = 23
a,
Đặt \(\frac{x}{2}=\frac{y}{3}=k\Rightarrow x=2k,y=3k\)
=> xy = 2k3k = 6k2 = 54
=> k2 = 9
=> k = +-3
=> [x,y] = [-6;-9], [6;9]
b,
\(\frac{5}{x}=\frac{3}{y}\Leftrightarrow\frac{25}{x^2}=\frac{9}{y^2}\)
áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{25}{x^2}=\frac{9}{y^2}=\frac{25-9}{x^2-y^2}=\frac{16}{4}=4\)
\(\Rightarrow\hept{\begin{cases}x^2=\frac{25}{4}\Rightarrow x=\frac{5}{2}\\y^2=\frac{9}{4}\Rightarrow y=\frac{3}{2}\end{cases}}\)
c,
\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\)
\(\Rightarrow\frac{1+4y}{24}=\frac{1+6y}{6x}=\frac{1+2y}{18}=\frac{1+2y+1+6y}{18+6x}=\frac{2+8y}{18+6x}=\frac{2\left[1+4y\right]}{2\left[9+3x\right]}=\frac{1+4y}{9+3x}\)
=> 24 = 9 + 3x
=> 3x = 15
=> x = 5
\(\frac{1+2y}{18}=\frac{1+4y}{24}\Leftrightarrow24\left[1+2y\right]=18\left[1+4y\right]\Leftrightarrow24+48y=18+72y\)
=> 24 + 48y - 18 = 72y
=> 6 + 48y = 72y
=> 6 = 24y
=> y = 1/4
ta có A=x+\(\sqrt{x}+1\)=\(\left(\sqrt{x}\right)^2+\sqrt{x}+1\)
ta thấy \(\sqrt{x}^2+\sqrt{x}\ge0\)
=>\(\sqrt{x}^2+\sqrt{x}+1\ge1\)
do đó min A=1
dấu bằng xảy ra \(\Leftrightarrow\)\(\sqrt{x}^2=0\)
\(\Leftrightarrow\)x=0
vậy GTNNcủa A=1 tại x=0
Bài 1:
a) \(\left(3x-\frac{4}{5}\right)^2+\left(2y+\frac{3}{7}\right)^2=0\)
\(\Rightarrow\left\{\begin{matrix}3x-\frac{4}{5}=0\\2y+\frac{3}{7}=0\end{matrix}\right.\rightarrow\left\{\begin{matrix}3x=\frac{4}{5}\\2y=-\frac{3}{7}\end{matrix}\right.\rightarrow\left\{\begin{matrix}x=\frac{4}{15}\\y=-\frac{3}{14}\end{matrix}\right.\)
Bài 1:
\(A=\frac{a+b}{b+c}.\)
Ta có:
\(\frac{b}{a}=2\Rightarrow\frac{b}{2}=\frac{a}{1}\) (1)
\(\frac{c}{b}=3\Rightarrow\frac{c}{3}=\frac{b}{1}\) (2)
Từ (1) và (2) \(\Rightarrow\frac{b}{2}=\frac{c}{6}.\)
\(\Rightarrow\frac{a}{1}=\frac{b}{2}=\frac{c}{6}=\frac{a+b}{3}=\frac{b+c}{8}.\)
\(\Rightarrow A=\frac{a+b}{b+c}=\frac{3}{8}\)
Vậy \(A=\frac{a+b}{b+c}=\frac{3}{8}.\)
Bài 2:
a) \(\frac{72-x}{7}=\frac{x-40}{9}\)
\(\Rightarrow\left(72-x\right).9=\left(x-40\right).7\)
\(\Rightarrow648-9x=7x-280\)
\(\Rightarrow648+280=7x+9x\)
\(\Rightarrow928=16x\)
\(\Rightarrow x=928:16\)
\(\Rightarrow x=58\)
Vậy \(x=58.\)
b) \(\frac{x+4}{20}=\frac{5}{x+4}\)
\(\Rightarrow\left(x+4\right).\left(x+4\right)=5.20\)
\(\Rightarrow\left(x+4\right).\left(x+4\right)=100\)
\(\Rightarrow\left(x+4\right)^2=100\)
\(\Rightarrow x+4=\pm10.\)
\(\Rightarrow\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10-4\\x=\left(-10\right)-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)
Vậy \(x\in\left\{6;-14\right\}.\)
Chúc bạn học tốt!
Bài 2:
a, \(\frac{72-x}{7}=\frac{x-40}{9}\)
\(\Rightarrow\left(72-x\right).9=\left(x-40\right).7\)
\(\Rightarrow9.72-9.x=7.x-7.40\)
\(\Rightarrow648-9x=7x-280\)
\(\Rightarrow-9x-7x=-280-648\)
\(\Rightarrow-16x=-648\)
\(\Rightarrow x=58\)
Vậy \(x=58\)