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\(=-\dfrac{7}{2}x+1+\dfrac{5}{4}x-3-\dfrac{1}{2}x\left(2x^2+x-2x-1\right)\)
\(=\dfrac{-9}{4}x-2-x^3-\dfrac{1}{2}x^2+x^2+\dfrac{1}{2}x\)
\(=-x^3+\dfrac{1}{2}x^2-\dfrac{7}{4}x-2\)
a) \(f\left(x\right)=5x^3-7x^2+2x+5\)
\(\Rightarrow f\left(1\right)=5.1^3-7.1^2+2.1+5\)
\(\Rightarrow f\left(1\right)=5.1-7.1+2+5\)
\(\Rightarrow f\left(1\right)=5-7+7\)
\(\Rightarrow f\left(1\right)=5\)
Vậy f(1) = 5.
\(g\left(x\right)=7x^3-7x^2+2x+5\)
\(\Rightarrow g\left(\frac{1}{2}\right)=7.\left(\frac{1}{2}\right)^3-7.\left(\frac{1}{2}\right)^2+2.\frac{1}{2}+5\)
\(\Leftrightarrow g\left(\frac{1}{2}\right)=7.\frac{1}{8}-7.\frac{1}{4}+1+5\)
\(\Leftrightarrow g\left(\frac{1}{2}\right)=\frac{7}{8}-\frac{14}{8}+6\)
\(\Leftrightarrow g\left(\frac{1}{2}\right)=\frac{-7}{8}+\frac{48}{8}\)
\(\Leftrightarrow g\left(\frac{1}{2}\right)=\frac{41}{8}\)
Vậy \(g\left(\frac{1}{2}\right)=\frac{41}{8}\)
\(h\left(x\right)=2x^3+4x+1\)
\(\Rightarrow h\left(0\right)=2.0^3+4.0+1\)
\(\Rightarrow h\left(0\right)=0+0+1\)
\(\Rightarrow h\left(0\right)=1\)
Vậy \(h\left(0\right)=1\)
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
a) 2x - 5 = 3 + 2x - 7x
=> 2x - 2x + 7x = 3 +5
=> 7x = 8
=> x = 8/7
b) \(\left(2x-1\right)^2=\left(2x-1\right)^5\)
=> \(\left(2x-1\right)^2-\left(2x-1\right)^5=0\)
=> \(\left(2x-1\right)^2\left[1-\left(2x-1\right)^3\right]=0\)
=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\1-\left(2x-1\right)^3=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^3=1\end{cases}}\)
=> \(\orbr{\begin{cases}2x=1\\2x-1=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=2\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)
1) \(\left(x-5\right)\left(x+7\right)-7x\left(x+3\right)\)
\(=x^2+7x-5x-35-7x^2-21x\)
\(=-6x^2-19x-35\)
2) \(\left(x+5\right)\left(x+7\right)-\left(x-4\right)\left(x+3\right)\)
\(=x^2+5x+7x+35-\left(x^2+3x-4x-12\right)\)
\(=x^2+12x+35-x^2+x+12\)
\(=13x+47\)
3) \(\left(2x-3\right)\left(x+4\right)+\left(-x+1\right)\left(x-2\right)\)
\(=2x^2+8x-3x-12-x^2+2x+x-2\)
\(=x^2+8x-14\)
1. ( 2x-3 ) - ( x - 5 ) = 2x - 2017
=> 2x - 3 - x + 5 = 2x - 2017
=> x + 2 = 2x - 2017
=> -x = -2019
=> x = 2019
1.
(2x-3)-(x-5)=2x-2017
=>2x-3=2x-2017+x-5=3x-2022
=>x=2022-3=2019
2.
Ta thấy:
|x-5| >= 0 (với mọi giá trị x)
|x+2|>=0(với mọi giá trị x)
=>|x-5|+|x+2|>=0(với mọi giá trị x)
hay 7x >= 0 (với mọi giá trị x)=>x>=0
Do đó |x-5|+|x+2|=x-5+x+2
hay x-5+x+2=7x
=>2x-3=7x
=>-5x=3
=>x=-3/5