Tìm Min hoặc Max:
a/ \(B=\frac{2}{-x^2+6x-12}\)
b/ \(C=\frac{x^2+x+1}{\left(x+1\right)^2}\)
c/ \(D=\frac{x}{\left(x+1\right)^2}\)
d/ \(E=\frac{-x^2+2x-2016}{x^2}\)
Tìm Min của các biểu thức sau :
a) A= \(\frac{4x^2-6x+1}{\left(2x+1\right)^2}\)
b) B= \(\frac{x}{\left(x+10\right)^2}\)
c) C= \(\frac{x}{\left(x+2016\right)^2}\)
d) D= \(\frac{x^2-2x+2000}{x^2}\)
e) E= \(\frac{x^2-2x+2015}{2015x^2}\)
f) F= \(\frac{x}{\left(x+2000\right)^2}\)
\(595655225+455+963+852+741+9563+855282552\)=
tìm x
a) \(\frac{x-1}{2}+\frac{x-2}{5}=\frac{1}{4}+\frac{x-7}{10}\)
b) \(3-\frac{2}{2x-3}=\frac{2}{5}+\frac{1}{2x-3}-\frac{3}{2}\)
c)\(7\cdot\left(x-1\right)+2x\cdot\left(1-x\right)=0\)
d) \(\frac{x+1}{2008}+\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{x+10}{2009}+\frac{x+11}{2008}+\frac{x+12}{2007}\)
e) \(\frac{2}{\left(x-1\right)\cdot\left(x-3\right)}+\frac{5}{\left(x-3\right)\cdot\left(x-8\right)}+\frac{12}{\left(x-8\right)\cdot\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)
bài 1:giải các pt sau:
a/\(\frac{1-x}{x+1}\)+3=\(\frac{2x+3}{x+1}\)
b/\(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2+10}{2x-3}\)
c/\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
d/\(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
e/\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
f\(\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
Giải các PT sau :
a,\(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{\left(x+2\right)\left(x-2\right)}\)
b,(2 - 3x) (x + 11) = (3x - 2) (2 - 5x)
c,\(\frac{3x-2}{6}-5=\frac{3-2\left(x+7\right)}{4}\)
d,\(\frac{x}{x-3}+\frac{x}{x +1}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)
e,\(\frac{x+1}{5}+\frac{x+2}{4}=\frac{x+3}{3}+\frac{x+4}{2}\)
Tìm MIN
a) A = \(3\left|2x-1\right|-4\)
b) B = \(x^4+3\left|y-2\right|-5\)
c) C = \(\left(x-\frac{2}{7}\right)^{2016}+\left(0,2-\frac{1}{5}y\right)^{2014}+\left(-1\right)^{2015}\)
d) \(D=\left|x-3\right|+\left|x+\frac{3}{2}\right|\)
a) \(\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{x}+x-2\right)\)
b) \(\left(\frac{3x}{1-3x}+\frac{2x}{3x+1}\right):\frac{6x^2+10x}{1-6x+9x^2}\)
c) \(\frac{x^3}{x-1}-\frac{x^2}{x+1}-\frac{1}{x-1}+\frac{1}{x+1}\)
d) \(\frac{x^3+x^2-2x-20}{x^2-4}-\frac{5}{x+2}+\frac{3}{x-2}\)
e) \(\left[\frac{x^2-y^2}{xy}-\frac{1}{x+y}\left(\frac{x^2}{y}-\frac{y^2}{x}\right)\right]:\frac{x-y}{x}\)
Mọi người làm ơn giúp mình đi! Mình đang cần gấp lắm!
a,\(\frac{3}{x}+\frac{1}{x+3}+\frac{3}{x+6}+\frac{1}{x+7}=\frac{1}{1-x}\)
b, \(\frac{1}{x-5}+\frac{1}{x-2}+\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+3}=\frac{3x-3}{4}\)
c,\(\frac{1}{x-3}+\frac{1}{3x+1}+\frac{10x-13}{4x-6}=\frac{1}{x+1}+\frac{1}{2x-1}+\frac{1}{3x+7}\)
d,\(\frac{x^2+x+1}{2x-1}\left(\frac{3x^2-x+5}{4x-2}-3\right)=8\)
e,\(\frac{2x^2-3}{3x-1}\left(2x-\frac{7+4x}{3x-1}\right)=2\)
f,\(\frac{x\left(3x-1\right)\left(3x^2+1\right)\left(6x^2-3x-1\right)}{\left(x+1\right)^3}=\frac{1}{2}\)
g, \(x\left(x^2+2\right)\left(x^2+2x+8+\frac{12}{x-2}\right)=3\left(x-2\right)\)
xác định các số hữu tỉ a,b,c,d sao cho:
a,\(\frac{1}{x\left(x+1\right)\left(x+2\right)}=\frac{a}{x\left(x+1\right)}+\frac{b}{\left(x+1\right)\left(x+2\right)}\)
b,\(\frac{x^3}{x^4-1}=\frac{a}{x-1}+\frac{b}{x+1}+\frac{cx+d}{x^2+1}\)
c,\(\frac{2x^2-x+1}{\left(x+1\right)\left(x-2\right)^2}=\frac{a}{x+1}+\frac{b}{x-2}+\frac{c}{x-2}\)
1.Tính:
\(a,A=\sqrt{12\frac{1}{4}}.\left(\frac{-2}{7}\right)^2-\left[2,\left(4\right).2\frac{5}{11}\right]:\left(\frac{-42}{5}\right)\)
\(B=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{2016}{3^{2016}}\)
2. Tìm x,y,z biết:
a) \(\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|-6x=0\)
b) \(\sqrt{\left(x+\sqrt{5}\right)^2}+\sqrt{\left(y+\sqrt{3}\right)^2}+\left|x-y-z\right|=0\)
c) \(\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-3}{4}\) và x-2y+3z=14.
d) \(5^x+5^{x+1}+5^{x+2}=3875\).
3. a) Cho bốn số a,b,c,d>0 thỏa mãn: \(\frac{1}{c}=\frac{ }{1}2.\left(\frac{1}{b}+\frac{1}{a}\right)\)và b là trung bình cộng của a và c. Chứng minh rằng bốn số đó lập nên một tỉ lệ thức.
b) Cho tỉ lệ thức: \(\frac{2a+13b}{3a-7b}=\frac{2c+13d}{3c-7d}\) (với a,b,c,d khác 0)
Chứng minh rằng: \(\frac{a}{b}=\frac{c}{d}\)
Bài 2:
a) \(\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|-6x=0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=6x\)
Ta có: \(\left|x+1\right|\ge0;\left|x+2\right|\ge0;\left|x+4\right|\ge0;\left|x+5\right|\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|\ge0\)
\(\Rightarrow6x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=x+1+x+2+x+4+x+5=6x\)
\(\Rightarrow4x+12=6x\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\)
Vậy x = 6
b) Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-3}{4}=\frac{2y-6}{6}=\frac{3z-9}{12}=\frac{x-2-2y+6+3z-9}{2-6+12}=\frac{\left(x-2y+3z\right)-\left(2-6+9\right)}{8}\)
\(=\frac{14-5}{8}=\frac{9}{8}\)
+) \(\frac{x-2}{2}=\frac{9}{8}\Rightarrow x-2=\frac{9}{4}\Rightarrow x=\frac{17}{4}\)
+) \(\frac{y-3}{3}=\frac{9}{8}\Rightarrow y-3=\frac{27}{8}\Rightarrow y=\frac{51}{8}\)
+) \(\frac{z-3}{4}=\frac{9}{8}\Rightarrow z-3=\frac{9}{2}\Rightarrow z=\frac{15}{2}\)
Vậy ...
c) \(5^x+5^{x+1}+5^{x+2}=3875\)
\(\Rightarrow5^x+5^x.5+5^x.5^2=3875\)
\(\Rightarrow5^x.\left(1+5+5^2\right)=3875\)
\(\Rightarrow5^x.31=3875\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
Vậy x = 3