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a)\(f\left(-1\right)=\left(-1\right)^2+5\cdot\left(-1\right)=1+\left(-5\right)=-4\)
\(f\left(-2\right)=\left(-2\right)^2+5\cdot\left(-2\right)=4+\left(-10\right)=-6\)
\(f\left(0\right)=0^2+5\cdot0=0\)
b)\(f\left(x\right)=-6\Leftrightarrow x^2+5x=-6\)
\(x^2+5x-\left(-6\right)=0\)
\(x^2+5x+6=0\)
\(x^2+2x+3x+6=0\)
\(x\left(x+2\right)+3\left(x+2\right)=0\)
\(\left(x+2\right)\left(x+3\right)=0\)
\(\Rightarrow x+2=0\) hoặc x+3=0
\(\Rightarrow\)x=-2 hoặc -3
a) f(-1) = (-1)2 + 5(-1) = -4 =y
tuong tu
b) x2 + 5x = -6
x2 +5x +6 = 0 => x2 +3x +2x +6 = 0
(x+3)(x+2) = 0
x = -3; x = -2
( chiều yên tâm đi học r)
Bài 2:
f(x)=x^2; g(x)=2/x
f(g(x))=(2/x)^2=4/x^2
g(f(x))=g(x^2)=2/x^2
Câu c thôi nhé!
c) Ta có: \(f\left(x\right)=3x^2-1\left(1\right)\)
\(f\left(-x\right)=3\left(-x\right)^2-1=3x^2-1\left(2\right)\)
Từ 1 và 2 \(\Rightarrow f\left(x\right)=f\left(-x\right)\)
\(\rightarrowĐPCM.\)
1.a) Theo đề bài,ta có: \(f\left(-1\right)=1\Rightarrow-a+b=1\)
và \(f\left(1\right)=-1\Rightarrow a+b=-1\)
Cộng theo vế suy ra: \(2b=0\Rightarrow b=0\)
Khi đó: \(f\left(-1\right)=1=-a\Rightarrow a=-1\)
Suy ra \(ax+b=-x+b\)
Vậy ...
a)F(x)=8+(-5x+3x)+(6x2 -3x2)+3x3
=8-2x+3x2+3x3
G(x)=-6+(12x2-9x2)+3x3
=-6+3x2+3x3
b)P(x)=8-2x+3x2+3x3-6+3x2+3x3
=(8-6)-2x+(3x2+3x2)+(3x3+3x3)
=2-2x+6x2+6x3
d)_thay \(\dfrac{1}{3}\) vào biểu thức F(x) ta có:
8-2.\(\dfrac{1}{3}\)+\(3.\left(\dfrac{1}{3}\right)^2\)+3.\(\left(\dfrac{1}{3}\right)^3\)
8-\(\dfrac{2}{3}\)+3.\(\dfrac{1}{9}\)+3.\(\dfrac{1}{27}\)
8-\(\dfrac{2}{3}\)+\(\dfrac{3}{9}\)+\(\dfrac{3}{27}\)
8-\(\dfrac{10}{27}\)
\(\dfrac{206}{27}\)
biểu thức G(x) tương tự chỗ nào có x bạn thay thành \(-\dfrac{1}{3}\)và tính thôi
c)mình chịu
a: \(f\left(-1\right)=3-7=-4\)
\(f\left(\dfrac{1}{5}\right)=\dfrac{3}{25}-7=\dfrac{-172}{25}\)
b: f(x)=-20/3
\(\Leftrightarrow3x^2-7=-\dfrac{20}{3}\)
\(\Leftrightarrow3x^2=\dfrac{1}{3}\)
\(\Leftrightarrow x^2=\dfrac{1}{9}\)
=>x=1/3 hoặc x=-1/3
đề k rõ nhung t hiu, bn hoc toan tb, kha hay joi để t lam theo bn
y=f(x)=4x - 1
a) f(-1)=4.(-1) - 1= -5
f(1)=4.1 - 1=3
f(1/4)=4.(1/4) - 1 =0
b) Với f(x)=0 => x=( 0+1)/4=0,25
=> y= f(0,25)=4.(0,25) - 1
Ta có y = f(x) = 3x2 + 1. Do đó
f(\(\dfrac{1}{2}\)) = 3.\(\left(\dfrac{1}{2}\right)^2\) + 1 = \(\dfrac{3}{4}\)+ 1 = \(\dfrac{7}{4}\)
f(1) = 3.12 + 1 = 3.1 + 1 = 3 + 1 = 4
f(3) = 3.32 + 1 = 3.9 + 1 = 27 + 1 = 28.
Ta có:\(f\left(x\right)=0\Rightarrow|3x-1|=0\Rightarrow3x-1=0\)
\(3x=0+1=1\)
\(x=1:3=\dfrac{1}{3}\)
\(f\left(x\right)=1\Rightarrow|3x-1|=1\Rightarrow3x-1=\pm1\)
*Với \(3x-1=1\Rightarrow3x=1+1=2\)
\(x=2:3=\dfrac{2}{3}\)
*Với \(3x-1=-1\Rightarrow3x=-1+1=0\)
\(x=0:3=0\)
\(f\left(x\right)=\dfrac{1}{2}\Rightarrow|3x-1|=\dfrac{1}{2}\Rightarrow3x-1=\pm\dfrac{1}{2}\)
*Với \(3x-1=\dfrac{1}{2}\Rightarrow3x=\dfrac{1}{2}+1=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}:3=\dfrac{3}{2}.\dfrac{1}{3}=\dfrac{1}{2}\)
*Với \(3x-1=-\dfrac{1}{2}\Rightarrow3x=-\dfrac{1}{2}+1=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}:3=\dfrac{1}{2}.\dfrac{1}{3}=\dfrac{1}{6}\)
\(f\left(x\right)=-\dfrac{2010}{2011}\Rightarrow|3x-1|=-\dfrac{2010}{2011}\Rightarrow x\in\varnothing\)
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