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a) \(f\left(x\right)-g\left(x\right)=\left[x\left(x^2-2x+7\right)-1\right]-\left[x\left(x^2-2x-1\right)-1\right]\)
\(f\left(x\right)-g\left(x\right)=x^3-2x^2+7x-1-x^3+2x^2+x+1\)
\(f\left(x\right)-g\left(x\right)=8x\)
\(f\left(x\right)+g\left(x\right)=x\left(x^2-2x+7\right)-1+x\left(x^2-2x-1\right)-1\)
\(f\left(x\right)+g\left(x\right)=x^3-2x^2+7x-1+x^3-2x^2-x-1\)
\(f\left(x\right)+g\left(x\right)=2x^3-4x^2+6x-2\)
b) 8x=0
=> x=0
=> Nghiệm đa thức f(x)-g(x)
c) Thay \(x=-\frac{3}{2}\)vào BT f(x)+g(x) ta được :
\(2.\left(-\frac{3}{2}\right)^3-4\left(-\frac{3}{2}\right)^2+6\left(-\frac{3}{2}\right)-2\)
\(=6,75+9-9-2\)
\(=4,75\)
#H
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\(a,\frac{-3}{2}-2x+\frac{3}{4}=-1\)
\(\frac{-3}{2}-2x=-1-\frac{3}{4}\)
\(\frac{-3}{2}-2x=\frac{-7}{4}\)
\(2x=\frac{-7}{4}+\frac{-3}{2}\)
\(2x=\frac{-13}{4}\)
\(x=\frac{-13}{4}:2\)
\(x=\frac{-13}{4}.\frac{1}{2}\)
\(x=\frac{-13}{8}\)
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\(3^{-2}.2^x+2^x.3=\dfrac{7}{36}\)
\(=>2^x\left(\dfrac{1}{9}+3\right)=\dfrac{7}{36}\)
\(=>2^x.\dfrac{28}{9}=\dfrac{7}{36}\)
\(=>2^x=\dfrac{1}{16}\)
\(=>2^x=2^{-4}\)
\(=>x=-4\)
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7x = 3y
=> x/3 = y/7
áp dụng tc của dãy tỉ số = nhau ta có :
x/3 = y/7 = (x-y)/(3-7) mà x - y = 16
=> x/3 = y/7 = -4
=> x = -12 và y = -28
7.x=3.y
\(\Leftrightarrow\)x/3=y/7
Áp dụng...........:
x/3=y/7=x-y/3-7=\(-\frac{16}{4}\) =-4
x/3=-4
x=-12
y/7=-4
y=-28
Vậy...
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\(\left|x+\dfrac{1}{7}\right|-\dfrac{2}{3}=0\)
\(\Rightarrow\left|x+\dfrac{1}{7}\right|=0+\dfrac{2}{3}\\ \Rightarrow\left|x+\dfrac{1}{7}\right|=\dfrac{2}{3}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{7}=\dfrac{2}{3}\\x+\dfrac{1}{7}=-\dfrac{2}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}-\dfrac{1}{7}\\x=-\dfrac{2}{3}-\dfrac{1}{7}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{21}\\x=-\dfrac{17}{21}\end{matrix}\right.\)
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x/y=3/4
=>x/3=y/4
=>x/15=y/20
y/z=5/7
=>y/5=z/7
=>y/20=z/28
=>x/15=y/20=z/28=(2x+3y-z)/(2*15+3*20-28)=186/62=3
=>x=45; y=60; z=84
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Cái này khá ez :>>
\(a,A\left(x\right)+\left(3x^2-4x+1\right)=5x-x^2\)
\(A\left(x\right)=5x-x^2-3x^2+4x-1\)
Ta có : \(9x-4x^2-1=0\)
Vậy phương trình vô nghiệm.
b, \(A\left(x\right)=5x^3-2x=x^3+x-1\)
\(A\left(x\right)=x^3+x-1-5x^3+2x\)
Ta có : \(-4x^3+3x-1=0\)
\(\left(-4x^2-4x+1\right)\left(x+1\right)=0\)
\(\left(2x-1\right)^2\left(x+1\right)=0\)
\(\orbr{\begin{cases}\left(2x-1\right)^2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-1\end{cases}}}\)