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\(1.\left(x-5\right)\left(x+5\right)-\left(x+4\right)^2+\left(4x+1\right)^3=\left(4x+2\right)\left(16x^2-8x+4\right)+12x\left(4x-1\right)\)⇔ \(x^2-25-x^2-8x-16+64x^3+48x^2+12x+1=64x^3+8+48x^2-12x\)⇔ \(16x-48=0\)
⇔ \(x=3\)
KL..........
1) Ta có: \(\left(x+5\right)\left(x+2\right)-3\left(4x-3\right)=\left(5-x\right)^2\)
\(\Leftrightarrow x^2+2x+5x+10-12x+9=25-10x+x^2\)
\(\Leftrightarrow x^2-5x+19-25+10x-x^2=0\)
\(\Leftrightarrow5x-6=0\)
\(\Leftrightarrow5x=6\)
\(\Leftrightarrow x=\frac{6}{5}\)
Vậy: \(x=\frac{6}{5}\)
2) Ta có: \(\left(x+2\right)^3-\left(x-2\right)^3=12x\left(x-1\right)-8\)
\(\Leftrightarrow x^3+6x^2+12x+8-\left(x^3-6x^2+12x-8\right)=12x^2-12x-8\)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8-12x^2+12x+8=0\)
\(\Leftrightarrow12x+24=0\)
\(\Leftrightarrow12x=-24\)
\(\Leftrightarrow x=-2\)
Vậy: x=-2
3) Ta có: \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2+27x-30=0\)
\(\Leftrightarrow15x-30=0\)
\(\Leftrightarrow15x=30\)
\(\Leftrightarrow x=2\)
Vậy: x=2
4) Ta có: \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(\Leftrightarrow48x^2-12x-20x+5+3x-48x^2-7+112x-81=0\)
\(\Leftrightarrow83x-83=0\)
\(\Leftrightarrow83x=83\)
\(\Leftrightarrow x=1\)
Vậy: x=1
bài 2
P= (x+1)(x2-x+1)+x-(x-1)(x2+x+1)+2010 với x = -2010
= (x3+1) + x - (x3-1) + 2010
= x3 + 1 + x - x3 + 1 + 2010
= x + 2 + 2010
= 2010 + 2 + 2010
=4022
Q=16x(4x2-5)-(4x+1)(16x2-4x + 1) với x = 1/5
= (4x)3-16.5x - [(4x)3+1]
= (4x)3 - 16.5x - (4x)3 - 1
= -16.5x - 1
= -16.5.1/5 - 1
= -16-1
=-17
a) (x-3)(x2+3x+9)-x(x-4)(x+4)=41
<=> x3 - 33 - x(x2 - 42) = 41
<=> x3 - 27 - x3 + 16x = 41
<=> 16x = 68
<=> x= 4,25
b) (x+2)(x2-2x+4)-x(x2+2)=4
<=> x3 + 23 - x3 - 2x =4
<=> 8 - 2x = 4
<=> 2x = 4
<=> x= 1/2
2.
A = x2 - 4x + 10 = (x2 - 2.x.2 + 22) + 6 = (x - 2)2 + 6 \(\ge\) 6
( do (x - 2)2 \(\ge\) 0)
Vậy: GTNN của A là 6 (tại x = 2)
B = x2 - x + 1 = (x2 - 2.x.\(\frac{1}{2}\) + \(\frac{1}{4}\)) + \(\frac{3}{4}\) = \(\left(x-\frac{1}{2}\right)^2\) + \(\frac{3}{4}\) \(\ge\) \(\frac{3}{4}\)
Vậy: GTNN của B là \(\frac{3}{4}\) (tại x = \(\frac{1}{2}\) )
C = 2x2 - 8x = 2 (x2 - 4x) = 2(x2 - 2.x.2 + 4) - 8 = 2(x - 2)2 - 8 \(\ge\) -8
Vậy : GTNN của C là -8 (tại x = 2)
Bài 1:
a)
\((x-5)(2x-1)-4x(x+2)=-(x-1)^2-2x(x-3)\)
\(\Leftrightarrow (2x^2-11x+5)-(4x^2+8x)=-(x^2-2x+1)-(2x^2-6x)\)
\(\Leftrightarrow -2x^2-19x+5=-3x^2+8x-1\)
\(\Leftrightarrow x^2-27x+6=0\)
\(\Leftrightarrow (x-\frac{27}{2})^2=\frac{705}{4}\Rightarrow \left[\begin{matrix} x-\frac{27}{2}=\frac{\sqrt{705}}{2}\\ x-\frac{27}{2}=\frac{-\sqrt{705}}{2}\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=\frac{27+\sqrt{705}}{2}\\ x=\frac{27-\sqrt{705}}{2}\end{matrix}\right.\)
b)
\((4x-1)-(2x+3)^2-12x(x+3)=1\)
\(\Leftrightarrow 4x-1-(4x^2+12x+9)-(12x^2+36x)=1\)
\(\Leftrightarrow -16x^2-44x-11=0\)
\(\Leftrightarrow 16x^2+44x+11=0\)
\(\Leftrightarrow (4x+\frac{11}{2})^2=\frac{77}{4}\)
\(\Rightarrow \left[\begin{matrix} 4x+\frac{11}{2}=\frac{\sqrt{77}}{2}\\ 4x+\frac{11}{2}=\frac{-\sqrt{77}}{2}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{\sqrt{77}-11}{8}\\ x=\frac{-\sqrt{77}-11}{8}\end{matrix}\right.\)
\(12x\left(x-4\right)-\left(x+1\right)\left(x^2-x+1\right)+\left(x-4\right)^3=\left(x-3\right)\left(x+1\right)-\left(x+5\right)^2\) ⇔ \(12x^2-48x-x^3-1+x^3-12x^2+48x-64=x^2-2x-3-x^2-10x-25\) ⇔ \(12x-37=0\)
⇔ \(x=\dfrac{37}{12}\)
Vậy ,....
1: \(\Leftrightarrow x^2-25-x^2-8x-16+\left(4x+1\right)^3=64x^3+8+48x^2-12x\)
\(\Leftrightarrow-8x-41+64x^3+48x^2+12x+1=64x^3+48x^2-12x+8\)
=>4x-40=-12x+8
=>16x=48
hay x=3
2: \(\Leftrightarrow12x^2-48x-x^3+1+x^3-12x^2+48x-64=x^2-2x-3-x^2-10x-25\)
\(\Leftrightarrow-63=-12x-28\)
=>12x+28=63
=>12x=35
hay x=35/12