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Bài 1:
Để \(F\left(x\right)=G\left(x\right)\) thì \(3x^2-8x+4=3x+4\)
\(\Leftrightarrow3x^2-11x=0\)
\(\Leftrightarrow x\left(3x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{11}{3}\end{matrix}\right.\)
a: f(a)=g(a)
=>5a-3=-1/2a+1
=>5,5a=4
=>\(a=\dfrac{4}{5.5}=\dfrac{8}{11}\)
b: f(b-2)=g(2b+4)
=>\(5\left(b-2\right)-3=-\dfrac{1}{2}\left(2b+4\right)+1\)
=>\(5b-13=-b-2+1=-b-1\)
=>6b=12
=>b=2
f(a) = g(a)
⇔ 5a - 3 = -a/2 + 1
⇔ 5a + a/2 = 1 + 3
⇔ 11a/2 = 4
⇔ 11a = 8
⇔ a = 8/11
Vậy a = 8/11 thì f(a) = g(a)
b) f(b - 2) = g(2b + 4)
⇔ 5.(b - 2) - 3 = -(2b + 4)/2 + 1
⇔ 5b - 10 - 3 = -b - 2 + 1
⇔ 5b + b = 1 + 13
⇔ 6b = 14
⇔ b = 7/3
Vậy b = 7/3 thì f(b - 2) = g(2b + 4)
\(a,f\left(-3\right)=9;f\left(-\dfrac{1}{2}\right)=\dfrac{1}{4};f\left(0\right)=0\\ g\left(1\right)=2;g\left(2\right)=1;g\left(3\right)=0\\ b,2f\left(a\right)=g\left(a\right)\\ \Leftrightarrow2a^2=3-a\\ \Leftrightarrow2a^2+a-3=0\\ \Leftrightarrow2a^2-2a+3a-3=0\\ \Leftrightarrow2a\left(a-1\right)+3\left(a-1\right)=0\\ \Leftrightarrow\left(2a+3\right)\left(a-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=1\\a=-\dfrac{3}{2}\end{matrix}\right.\)
b: Ta có: \(2\cdot f\left(a\right)=g\left(a\right)\)
\(\Leftrightarrow2a^2=3-a\)
\(\Leftrightarrow2a^2+a-3=0\)
\(\Leftrightarrow2a^2+3a-2a-3=0\)
\(\Leftrightarrow\left(2a+3\right)\left(a-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a=-\dfrac{3}{2}\end{matrix}\right.\)
a: f(-2)-g(1/2)
\(=5\left(-2\right)-3+4\cdot\dfrac{1}{2}-1\)
\(=-10-4+2=-10-2=-12\)
b: \(2\cdot f^2\left(-3\right)-3\cdot g^2\left(-2\right)\)
\(=2\cdot\left[5\cdot\left(-3\right)-3\right]^2-3\cdot\left[\left(-4\right)\left(-2\right)+1\right]^2\)
\(=2\cdot\left(-18\right)^2-3\cdot9^2\)
\(=648-3\cdot81=405\)