Bài 5: Hàm số

a, Ta có :

\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)...........\left(\dfrac{1}{10}-1\right)\)

\(=\left(\dfrac{1}{2}-\dfrac{2}{2}\right)\left(\dfrac{1}{3}-\dfrac{3}{3}\right).........\left(\dfrac{1}{10}-\dfrac{10}{10}\right)\)

\(=\dfrac{-1}{2}.\dfrac{-2}{3}...............\dfrac{-9}{10}\)

\(=\dfrac{-1.\left(-2\right)............\left(-9\right)}{2.3........9.10}\)

\(=\dfrac{-1}{10}< \dfrac{-1}{9}\)

\(\Leftrightarrow A< \dfrac{-1}{9}\)

b, \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)..........\left(\dfrac{1}{100}-1\right)\)

\(=\left(\dfrac{1}{4}-\dfrac{4}{4}\right)\left(\dfrac{1}{9}-\dfrac{9}{9}\right).........\left(\dfrac{1}{100}-\dfrac{100}{100}\right)\)

\(=\dfrac{-3}{4}.\dfrac{-8}{9}..............\dfrac{-99}{100}\)

\(=\dfrac{1.\left(-3\right).2\left(-4\right)............9\left(-11\right)}{2^2.3^2.......10^2}\)

\(=\dfrac{1.2.3........9}{2.3.......10}.\dfrac{\left(-3\right)\left(-4\right)....\left(-11\right)}{2.3...10}\)

\(=\dfrac{1}{10}.\dfrac{-11}{1}\)

\(=\dfrac{-11}{10}>\dfrac{-11}{21}\)

\(\Leftrightarrow B>\dfrac{-11}{21}\)

Ta có:

\(f\left(x\right)=-1\Rightarrow-2x=-1\Rightarrow x=\dfrac{1}{2}\)

\(f\left(x\right)=-\dfrac{2}{3}\Rightarrow-2x=-\dfrac{2}{3}\Rightarrow x=\dfrac{1}{3}\)

Vậy.................

Chúc bạn học tốt!!!

a) Xác đinh \(f\left(2\right);f\left(3\right)\)

\(f\left(x\right)=x^2-5x+6\)
\(f\left(x\right)=x^2-2x-3x+6\)

\(f\left(x\right)=x\left(x-2\right)-3\left(x-2\right)\)
\(f\left(x\right)=\left(x-2\right)\left(x-3\right)\)
\(f\left(2\right)=f\left(3\right)=0\)

b) Tìm x f(x) =6

\(f\left(x\right)=6\Rightarrow x^2-5x+6=6\)
\(\Leftrightarrow x^2-5x=0\)

\(\Rightarrow x\left(x-5\right)=0\)
\(\Rightarrow x=0\\ x=5\)

Bài1:

\(M=\dfrac{9-x}{4-x}=1+\dfrac{5}{4-x}\)

Để M đạt giá trị lớn nhất thì 4-x phải đặt giá trị nhỏ nhất

=>4-x đạt giá trị là số nguyên dương nhỏ nhất có thể

=>4-x=1

=>x=3

Thay x=3 vào M,ta có:

\(M=\dfrac{9-3}{4-3}=\dfrac{6}{1}=6\)

Vậy....

Bài2:

\(\left(x-2\right)^2+\left(2y-1\right)^2\)

Với mọi x;y thì \(\left(x-2\right)^2>=0;\left(2y-1\right)^2>=0\)

=>\(\left(x-2\right)^2+\left(2y-1\right)^2>=0\)

Để \(\left(x-2\right)^2+\left(2y-1\right)^2=0\) thì

\(\left(x-2\right)^2=0\) và \(\left(2y-1\right)^2=0\)

=>\(x-2=0\) và \(2y-1=0\)

=>\(x=2vay=\dfrac{1}{2}\)

Vậy....

\(M=\dfrac{9-x}{4-x}=\dfrac{5+4-x}{4-x}=\dfrac{5}{4-x}+\dfrac{4-x}{4-x}=\dfrac{5}{4-x}+1\)Để \(max_M\) thì \(\dfrac{5}{x-4}\) phải là số nguyên lớn nhất có thể

Vậy \(\dfrac{5}{x-4}=5\Rightarrow x=3\)

Thay vào biểu thức:

\(max_M=\dfrac{9-3}{4-3}=6\)

\(\left(x-2\right)^2+\left(2y-1\right)^2=0\)

\(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left(2y-1\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x-2\right)^2+\left(2y-1\right)^2\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(2y-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)

Ta có: \(3\left(x+1\right)^2+2\ge2\Rightarrow B\le\dfrac{8}{2}=4\)

Dấu " = " khi \(3\left(x+1\right)^2=0\Leftrightarrow x=-1\)

Vậy \(MAX_B=4\) khi x = -1

Ta có:

\(3\left(x+1\right)^2+2\ge2\)

\(\Rightarrow B\le\dfrac{8}{2}=4\)

Dấu '' = '' chỉ xảy ra khi

\(3\left(x+1\right)^2=0\Rightarrow x=-1\)

Vậy \(MAX_B=4\) khi \(x=-1\)

Chúc bạn học tốt!

\(B=\dfrac{8}{3\left(x+1\right)^2+2}\)

Với mọi x thì \(3\left(x+1\right)^2\ge0\Rightarrow3\left(x+1\right)^2+2\ge2\)

\(\Rightarrow B=\dfrac{8}{3\left(x+1\right)^2+2}\le4\)

Để B=4 thi:\(3\left(x+1\right)^2=0\)

\(\Rightarrow\left(x+1\right)^2=0\Rightarrow x+1=0\Rightarrow x=-1\)

Vậy....

\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\left(1\right)\\ \Leftrightarrow1+\dfrac{a}{b}+\dfrac{b}{a}+1-4\ge0\\ \Leftrightarrow\dfrac{a}{b}+\dfrac{b}{a}-2\ge0\left(2\right)\)

Áp dụng t/c \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\) nên (2) luôn đúng.Do đó:(1) đúng

Vậy...(đpcm)

\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\) ( đúng)

\(\Rightarrow1+\dfrac{a}{b}+\dfrac{b}{a}+1-4\ge0\)

\(\Rightarrow\dfrac{a}{b}+\dfrac{b}{a}-2\ge0\) ( đúng)

Ta áp dụng tính chất \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\)

Vậy .....

Chúc bạn học tốt!

\(a)2:\left(\dfrac{1}{2}-\dfrac{2}{3}\right)^3=2:\left(\dfrac{-1}{6}\right)^3=2:\dfrac{-1}{216}=2.\left(-216\right)=-432\)

\(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)

\(=\left(\dfrac{1}{3}-\dfrac{1}{4}\right).\left(\dfrac{1}{20}\right)^2\)

\(=\dfrac{1}{12}.\dfrac{1}{400}=\dfrac{1}{4800}\)

chúc bạn học tốt

Sửa đề: f(x)=1/3x

a: f(2/3)=2/9

f(-1)=-1/3

b: f(x)=-3

nên 1/3x=-3

hay x=-9

c: Khi x=-3 thì \(y=\dfrac{1}{3}\cdot\left(-3\right)=-1< >y_M\)

=>M không thuộc đồ thị

Khi x=6 thì \(y=\dfrac{1}{3}\cdot6=2=y_N\)

=>N thuộc đồ thị

Khi x=9 thì \(y=\dfrac{1}{3}\cdot9=3< >-3=y_P\)

=>P không thuộc đồ thị