- Cần luôn !!
Cho tam giác ABC . I giao điểm của 3 tia phân giác . IM \(\perp\)BC ; IK \(\perp\)AC ; IN \(\perp\)BC . Qua điểm A vẽ a // MN , b // NK . a cắt NA tại E , b cắt MN tại B ; ED cắt AC , AB tại P , Q . C/m : PQ//BC
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a)
\(x^2+x+\frac{1}{4}=4x^2\)
\(x^2+2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2=\left(2x\right)^2\)
\(\left(x+\frac{1}{2}\right)^2=\left(2x\right)^2\)
\(\Leftrightarrow x+\frac{1}{2}=2x\)
\(\Leftrightarrow x=\frac{1}{2}\)
2)
\(3x^2+6x+100\)
\(=3\left(x^2+2x+\frac{100}{3}\right)\)
\(=3\left(x^2+2\cdot x\cdot1+1^2+\frac{100}{3}\right)\)
\(=3\left[\left(x+1\right)^2+\frac{100}{3}\right]\)
\(=3\left(x+1\right)^2+100\ge100\forall x\left(đpcm\right)\)
a2 - a = a ( a - 1 )
mà a và a-1 là 2 số liên tiếp
=> 1 trong 2 số là số chẵn
=> a ( a - 1 ) chia hết cho 2 hay a2 - a chia hết cho 2
Ta có : \(a^2-a=a\left(a-1\right)\)
Vì \(a\left(a-1\right)\)là tích 2 số nguyên liên tiếp nên
\(a\left(a-1\right)⋮2\)
+ \(a^3-a=a\left(a^2-1\right)=a\left(a-1\right)\left(a+1\right)\)
Vì \(a\left(a-1\right)\left(a+1\right)\)là tích 3 số nguyên liên tiếp nên :
\(a\left(a-1\right)\left(a+1\right)⋮3\)
+ \(a^5-a=a\left(a^4-1\right)\)
\(=a\left(a^2-1\right)\left(a^2+1\right)\)
\(=\left(a-1\right)a\left(a+1\right)\left(a^2-4+5\right)\)
\(=\left(a-1\right)a\left(a+1\right)\left(a^2-4\right)+5\left(a-1\right)a\left(a+1\right)\)
\(=\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)\)\(+5\left(a-1\right)a\left(a+1\right)\)
Vì \(\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)\)là tích 5 số nguyên liên tiếp
\(\Rightarrow\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)⋮5\)
\(\Rightarrow a^5-a⋮5\)
x3-7x+6=x3-x-6x+6=x(x-1)(x+1)-6(x-1)=(x-1)(x2+x-6)=(x-1)(x-2)(x+3)
\(x^3-7x+6=x^3+3x^2-3x^2-9x+2x+6\)
\(=x^2.\left(x+3\right)-3x.\left(x+3\right)+2.\left(x+3\right)\)
\(=\left(x+3\right).\left(x^2-3x+2\right)\)
\(=\left(x+3\right).\left(x^2-x-2x+2\right)\)
\(=\left(x+3\right).\left[x.\left(x-1\right)-2.\left(x-1\right)\right]\)
\(=\left(x+3\right).\left(x-1\right).\left(x-2\right)\)
Đặt \(a=\frac{1}{315}\), \(b=\frac{1}{651}\)ta có :
\(A=\left(2+a\right)\cdot b-3a\left(3+1-b\right)-4ab+12a\)
\(\Rightarrow A=2b+ab-12a+3ab-4ab+12a\)
\(\Rightarrow A=2b=\frac{2}{651}\)
\(x^3-9x^2+6x+16=x^3-8x^2-x^2+8x-2x+16\)
\(=x^2.\left(x-8\right)-x.\left(x-8\right)-2.\left(x-8\right)\)
\(=\left(x-8\right).\left(x^2-x-2\right)=\left(x-8\right).\left(x^2-2x+x-2\right)\)
\(=\left(x+8\right)\left[x.\left(x-2\right)+\left(x-2\right)\right]\)
\(=\left(x+8\right).\left(x-2\right)\left(x+1\right)\)
\(x^3-9x^2+6x+16\)
\(=\left(x^3+x^2\right)-\left(10x^2+10x\right)+\left(16x+16\right)\)
\(=x^2.\left(x+1\right)-10x\left(x+1\right)+16\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-10x+16\right)\)
Tham khảo nhé~
Ta có \(2018-5x^2-y^2-4xy+x\)
\(=2018+\frac{1}{4}-\frac{1}{4}-4x^2-x^2-y^2-2.2xy+x\)
\(=2018+\frac{1}{4}-\left(2x\right)^2-2.2xy-y^2-x^2+2.\frac{1}{2}x-\frac{1}{4}\)
\(=\frac{8073}{4}-\left[\left(2x\right)^2+2.2xy+y^2\right]-\left[x^2-2.\frac{1}{2}x+\left(\frac{1}{2}\right)^2\right]\)
\(=\frac{8073}{4}-\left(2x+y\right)^2-\left(x-\frac{1}{2}\right)^2\ge\frac{8073}{4}\)( Vì \(\left(2x+y\right)^2\ge0;\left(x-\frac{1}{2}\right)^2\ge0\))
Dấu ''='' xảy ra khi \(\hept{\begin{cases}2x+y=0\\x-\frac{1}{2}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=-2x\\x=\frac{1}{2}\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-1\end{cases}}}\)
Vậy GTNN của \(2018-5x^2-y^2-4xy+x\)là \(\frac{8073}{4}\)khi \(x=\frac{1}{2}\)và\(y=-1\)
Ta có : \(a^7-a=a\left(a^6-a\right)\) \(=a\left(a^3-1\right)\left(a^3+1\right)\)
TH1 : \(a⋮7\)\(\Rightarrow a^7-a=a\left(a^3-1\right)\left(a^3+1\right)\)\(⋮\)\(7\)
TH2 : \(a=7k+1\)\(\left(k\inℤ\right)\)
\(\Rightarrow a^3-1⋮7\)\(\Rightarrow a^7-a=a\left(a^3-1\right)\left(a^3+1\right)\)\(⋮\)\(7\)
TH3 : \(a=7k+2\)\(\left(k\inℤ\right)\)
\(\Rightarrow a^3-1⋮7\)
\(\Rightarrow a^7-a=a\left(a^3+1\right)\left(a^3-1\right)\)\(⋮\)\(7\)
TH4 : \(a=7k+3\)\(\left(k\inℤ\right)\)
\(\Rightarrow a^3+1⋮7\)\(\Rightarrow a^7-a=a\left(a^3-1\right)\left(a^3+1\right)⋮7\)
TH5 : \(a=7k+4\)\(\left(k\inℤ\right)\)
\(\Rightarrow a^3-1⋮7\)\(\Rightarrow a^7-a=a\left(a^3-1\right)\left(a^3+1\right)⋮7\)
TH6 : \(a=7k+5\)\(\left(k\inℤ\right)\)
\(\Rightarrow a^3+1⋮7\)\(\Rightarrow a^7-a=a\left(a^3-1\right)\left(a^3+1\right)⋮7\)
TH7 : \(a=7k+6\)\(\left(k\inℤ\right)\)
\(\Rightarrow a^3+1⋮7\)\(\Rightarrow a^7-a=a\left(a^3-1\right)\left(a^3+1\right)⋮7\)
Vậy trong tất cả các trường hợp ta đều có : \(a^7-a⋮7\)
ta co: \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}.\)
\(\Rightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=0\)
=> x + y + z = 0
Lai co: x3 + y3 +z3 - 3xyz = (x+y+z).(x2+y2+z2 - xy - yz - zx)
x3 + y3 + z3 - 3xyz = 0
=> x3 + y3 + z3 = 3xyz
ta co: \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}.\)
=> 1/xy + 1/yz + 1/xz = 0
=> x + y + z = 0
Lai co: x3 + y3 +z3 - 3xyz = (x+y+z).(x2+y2+z2 - xy - yz - zx)
x3 + y3 + z3 - 3xyz = 0
=> x3 + y3 + z3 = 3xyz