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\(a,f\left(-\dfrac{1}{2}\right)=\dfrac{1}{4}+4=\dfrac{17}{4}\\ f\left(5\right)=25+4=29\\ b,f\left(x\right)=10=x^2+4\Leftrightarrow x^2=6\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{6}\\x=-\sqrt{6}\end{matrix}\right.\)
a/ Thay x =0 vào hàm số f(x) = 2x2 - 10 ta có
f(0) = 2 . 0 - 10 = -10
Thay x = 1 vào hàm số f(x) = 2x2 - 10 ta có
f(1) = 2 . 12 - 10 = 2 - 10 = -8
Thay \(x=-1\dfrac{1}{2}=-\dfrac{3}{2}\)vào hàm số f(x) ta có
\(f\left(-1\dfrac{1}{2}\right)=2.\left(-\dfrac{3}{2}\right)^2-10=\dfrac{9}{2}-\dfrac{20}{2}=-\dfrac{11}{2}\)
b/ f(x) = -2
\(\Leftrightarrow2x^2=8\)
\(\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)
a) Cho hàm số y = f(x) = -2x + 3.
Ta có: f(-2)= -2.(-2)+3
= 4+3=7
Ta có: f(0)= -2.0+3
= 0+3=3
Ta có: f(\(\dfrac{-1}{2}\))= -2.(-\(\dfrac{1}{2}\))+3
=\(\dfrac{-2.\left(-1\right)}{2}\)+3
=\(\dfrac{2}{2}\)+3
= 1+3= 4
Vậy f(-2)=7;f(0)=3;f( \(\dfrac{-1}{2}\))=4
b) Cho hàm số y = f(x) = -2x + 3
mà f(x)=5
Suy ra: f(x) = -2x + 3=5
hay -2x + 3=5
-2x=5-3
-2x=2
x=2:(-2)
x= -1
Cho hàm số y = f(x) = -2x + 3
mà f(x)=1
Suy ra: f(x) = -2x + 3=1
hay -2x + 3=1
-2x=1-3
-2x= -2
x= -2:(-2)
x=1
Vậy f(x)=5 thì x= -1 và f(x) = 1 thì x=1.
Lời giải:
a.
$f(-2)=(-2)(-2)+3=7$
$f(0)=(-2).0+3=3$
$f(\frac{-1}{2})=(-2).\frac{-1}{2}+3=4$
b.
$f(x)=-2x+3=5$
$\Rightarrow -2x=2$
$\Rightarrow x=-1$
$f(x)=-2x+3=1$
$\Rightarrow -2x=1-3=-2$
$\Rightarrow x=1$
Cho hàm số y=f(x)= −3x.
Ta có f(\(\dfrac{-3}{2}\)) = -3. (\(\dfrac{-3}{2}\))
= \(\dfrac{-3.\left(-3\right)}{2}\)
=\(\dfrac{9}{2}\)
Ta có f(-1) = -3. (-1)
= 3
Vậy f(\(\dfrac{-3}{2}\)) = \(\dfrac{9}{2}\) và f(-1) = 3.
\(f\left(-1\right)=2\Rightarrow-a+b-c+d=2\\ f\left(0\right)=1\Rightarrow d=1\\ f\left(1\right)=7\Rightarrow a+b+c+d=7\\ f\left(\dfrac{1}{2}\right)=3\Rightarrow\dfrac{1}{8}a+\dfrac{1}{4}b+\dfrac{1}{2}c+d=3\)
\(d=1\Rightarrow-a+b-c=1;a+b+c=6\\ \Rightarrow2b=7\\ \Rightarrow b=\dfrac{7}{2}\\ \Rightarrow\dfrac{1}{8}a+\dfrac{7}{8}+\dfrac{1}{2}c=2\\ \Rightarrow\dfrac{1}{2}\left(\dfrac{1}{4}a+\dfrac{7}{4}+c\right)=2\\ \Rightarrow\dfrac{1}{4}a+\dfrac{7}{4}+c=4\\ \Rightarrow a+7+4c=16\\ \Rightarrow a+4c=9;a+c=6-\dfrac{7}{2}=\dfrac{5}{2}\\ \Rightarrow3c=\dfrac{13}{2}\Rightarrow c=\dfrac{13}{6}\\ \Rightarrow a=\dfrac{5}{2}-\dfrac{13}{6}=\dfrac{1}{3}\)
Vậy \(\left(a;b;c;d\right)=\left(\dfrac{1}{3};\dfrac{7}{2};\dfrac{13}{6};1\right)\)
a: f(1)=1
=>\(a\cdot1^2+b\cdot1+1=1\)
=>a+b=0
f(-1)=3
=>\(a\cdot\left(-1\right)^2+b\cdot\left(-1\right)+1=3\)
=>a-b=2
mà a+b=0
nên \(a=\dfrac{2+0}{2}=1;b=2-1=1\)
b: a=1 và b=1 nên \(f\left(x\right)=x^2+x+1\)
\(\Leftrightarrow\dfrac{n}{f\left(n\right)}=\dfrac{n}{n^2+n+1}\)
Gọi d=ƯCLN(n^2+n+1;n)
=>\(\left\{{}\begin{matrix}n^2+n+1⋮d\\n⋮d\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}n^2+n+1⋮d\\n\left(n+1\right)⋮d\end{matrix}\right.\)
=>\(\left(n^2+n+1\right)-n\left(n+1\right)⋮d\)
=>\(1⋮d\)
=>d=1
=>ƯCLN(n^2+n+1;n)=1
=>\(\dfrac{n}{f\left(n\right)}=\dfrac{n}{n^2+n+1}\) là phân số tối giản
Bài 2:
x=13 nên x+1=14
\(f\left(x\right)=x^{14}-x^{13}\left(x+1\right)+x^{12}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+14\)
\(=x^{14}-x^{14}-x^{13}+x^{13}-...+x^3+x^2-x^2-x+14\)
=14-x=1
x=13 nên x+1=14
f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14
=x14−x14−x13+x13−...+x3+x2−x2−x+14=x14−x14−x13+x13−...+x3+x2−x2−x+14
=14-x=1
`e)3/(3x)-3/12=4/5-(7/x-2)`
`<=>1/x-1/4=4/5-7/x+2`
`<=>8/x=1/4+4/5+2=61/20`
`<=>1/x=61/160`
`<=>x=160/61`
`f)1/(x-1)+(-2)/3(3/4-6/5)=5/(2-2x)`
`<=>1/(x-1)+5/(2x-2)=2/3(3/4-6/5)=-3/10`
`<=>7/(2x-1)=-3/10`
`<=>2x-1=-70/3`
`<=>2x=-67/3`
`<=>x=-67/6`
(1)
a) x=\(\dfrac{-1}{12}-\dfrac{2}{3}\)=\(\dfrac{-3}{4}\)
b) 2x+1=3 => 2x=3-1=2 => x=1
(2)
f(2)=2.22+4=12
f(-1)=2.(-1)2+4=6
(1)
a) \(x+\dfrac{2}{3}=-\dfrac{1}{12}\\ \Rightarrow x=-\dfrac{1}{12}-\dfrac{2}{3}\\ \Rightarrow x=\dfrac{-1}{12}-\dfrac{8}{12}\\ \Rightarrow x=-\dfrac{9}{12}=-\dfrac{3}{4}\)
Vậy \(x=-\dfrac{3}{4}\)
b) \(\left(2x+1\right)^2=9\\ \Rightarrow\left(2x+1\right)^2=3^2=\left(-3\right)^2\\ \Rightarrow\left[{}\begin{matrix}2x+1=3\\2x+1=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=2\\2x=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{-2;1\right\}\)
(2)
\(y=f\left(x\right)=2x^2+4\\ f\left(2\right)=2\cdot2^2+4=8+4=12\\ f\left(-1\right)=2\cdot\left(-1\right)^2+4=2+4=6\)
Vậy \(f\left(2\right)=12\\ f\left(-1\right)=6\)