Bài 2:
a)Ta thấy: \(\left|\frac{1}{2}-x\right|\ge0\)
\(\Rightarrow A\ge0\)
Dấu "=" xảy ra khi \(x=\frac{1}{2}\)
Vậy \(Min_A=0\) khi \(x=\frac{1}{2}\)
b)Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(B=\left|\frac{1}{2}-x\right|+\left|x+2\right|\ge\left|\frac{1}{2}-x+x+2\right|=\frac{5}{2}\)
Dấu "=" xảy ra khi x=0
Vậy \(Min_B=\frac{5}{2}\) khi x=0
1. a, tính gt nhỏ nhất của biểu thức
A=\(\frac{2x^2-16x+41}{x^2-8x+22}\)
b, tính gt lớn nhất của biểu thúc
B=\(\frac{3x^2+9x+17}{3x^2+9x+7}\)
2. cho bt Q=\(\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right].\frac{4x^2+4x+1}{\left(x+3\right)\left(4-x\right)}\)
chứng minh rằng giá trị biểu thức sau ko hụ thuộc vào biến
a.\(\left(\frac{1}{3}+2x\right)\left(4x^2-\frac{2}{3}x+\frac{1}{9}\right)-\left(8x^3-\frac{1}{27}\right)\)
b.\(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3\left(1-x\right)x\)
c.\(y\left(x^2-y^2\right)\left(x^2+y^2\right)-y\left(x^4-y^4\right)\)
Bài 1 : dùng hẳng đẳng thức để khai triển và thu gọn
a) \(\left(2x^2+\frac{1}{3}\right)^3\)
b) \(\left(2x^2y-3xy\right)^3\)
c) \(\left(-3xy^4+\frac{1}{2}x^2y^2\right)^3\)
d) \(\left(-\frac{1}{3}ab^2-2a^3b\right)^3\)
e) \(\left(x+1\right)^3-\left(x-1\right)^3-6.\left(x-1\right).\left(x+1\right)\)
f) \(x.\left(x-1\right).\left(x+1\right)-\left(x+1\right).\left(x^2-x+1\right)\)
g) \(\left(x-1\right)^3-\left(x+2\right).\left(x^2-2x+4\right)+3.\left(x-4\right).\left(x+4\right)\)
h) \(3x^2.\left(x+1\right).\left(x-1\right)+\left(x^2-1\right)^3-\left(x^2-1\right).\left(x^4+x^2+1\right)\)
k) \(\left(x^4-3x^2+9\right).\left(x^2+3\right)+\left(3-x^2\right)^3-9x^2.\left(x^2-3\right)\)
l) \(\left(4x+6y\right).\left(4x^2-6xy+9y^2\right)-54y^3\)
cm các biểu thức sau ko phụ thuộc vào biến:
a,\(\left[\frac{2\left(x+1\right)\left(y+1\right)}{\left(x+1\right)^2-\left(y+1\right)^2}+\frac{x-y}{2x+2y+4}\right].\frac{2x+2}{x+y+2}+\frac{y+1}{y-x}\)
b,\(\left[2\left(x+y\right)+1-\frac{1}{1-2x-2y}\right]:\left[2x+2y-\frac{4x^2+8xy+4y^2}{2x+2y-1}\right]+2\left(x+y\right)\)
thực hiện phép tính
a,\(x^3+\left[\frac{x\left(2y^3-x^3\right)}{x^3+y^3}\right]^3-\left[\frac{y\left(2x^3-y^3\right)}{x^3+y^3}\right]^3\)
b,\(\frac{\frac{x\left(x+y\right)}{x-y}+\frac{x\left(x+z\right)}{x-z}}{1+\frac{\left(y-z\right)^2}{\left(x-y\right)\left(x-z\right)}}+\frac{\frac{y\left(y+z\right)}{y-z}+\frac{y\left(y+x\right)}{y-x}}{1+\frac{\left(z-x\right)^2}{\left(y-z\right)\left(y-x\right)}}+\frac{\frac{z\left(z+x\right)}{z-x}+\frac{z\left(z+y\right)}{z-y}}{1+\frac{\left(x-y\right)^2}{\left(z-x\right)\left(z-y\right)}}\)
c,\(\left[\frac{y+z-2x}{\frac{\left(y-z\right)^3}{y^3-z^3}+\frac{\left(x-y\right)\left(x-z\right)}{y^2+yz+z^2}}+\frac{z+x-2y}{\frac{\left(z-x\right)^3}{z^3-x^3}+\frac{\left(y-z\right)\left(y-x\right)}{z^2+xz+x^2}}+\frac{x+y-2z}{\frac{\left(x-y\right)^3}{x^3-y^3}+\frac{\left(z-x\right)\left(z-y\right)}{x^2+xy+y^2}}\right]:\frac{1}{x+y+z}\)
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,\(\frac{2x^2-x+1}{\left(x+1\right)\left(x-2\right)^2}=\frac{a}{x+1}+\frac{b}{x-2}+\frac{c}{x-2}\)
A=\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+9\right)\left(x+100\right)}\)
Giaỉ PT
a) \(\left(\frac{x-1}{99}+x-99\right)+\left(\frac{x-3}{97}+\frac{x-7}{93}\right)+\left(\frac{x-5}{95}+\frac{x-95}{5}\right)=6\)
b) \(\left(4x-5\right)^2\left(2x-3\right)\left(x-1\right)=9\)
c) \(\frac{5}{x-8}+1=\frac{23}{x^2-5x-24}-\frac{2}{x+3}\)
Help !!
Giúp tớ với
a)\(\frac{\left(x-4\right)^2}{7}-\frac{\left(x+4\right)^2}{5}\ge\frac{2\left(x^2-3\right)}{35}\)
b)\(\frac{\left(2x-1\right)^2}{2}-\frac{\left(2x+1\right)^2}{4}< \left(x-3\right)^2\)
c)\(|-5x+1|=6-3x\)
d)\(6\cdot|x+3|=-8x\)
e)\(7\cdot|x+1|=6-x\)