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a, f(x)=( x - 100 )( x5 - x4 + x3 - x2 + x ) - x + 25
=>f(100) = - 75
x3 - 100x2 - 101x + 1 tại x = 101
\(x^3-\left(101x-100x^2+1\right)x=101\)
\(x^2-\left(-9899x^2+1\right)x=101\)
\(x^2--9898x=101\)
\(x=101^2+9898\)
\(x=303\)
\(x^3-100x^2-101x+1\)
\(=x^3-101x^2+x^2-101x+1\)
\(=x^2\left(x-101\right)+x\left(x-101\right)+1\)
\(=101^2\left(101-101\right)+101\left(101-101\right)+1\)
\(=1\)
c)\(4x^4-101x^2+25=0\)
\(\Leftrightarrow4x^4-100x^2-x^2+25=0\)
\(\Leftrightarrow4x^2\left(x^2-25\right)-\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(4x^2-1\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+1=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{2}\\x=5\\x=-5\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{2};\frac{-1}{2};5;-5\right\}\)
a)\(\left(2x-1\right)^2=x+5\)
\(\Leftrightarrow4x^2-4x+1=x+5\)
\(\Leftrightarrow4x^2-5x-4=0\)
\(\Leftrightarrow4\left(x^2-\frac{5}{4}x-1\right)=0\)
\(\Leftrightarrow4\left(x^2-\frac{5}{4}x+\frac{25}{64}-\frac{89}{64}\right)=0\)
\(\Leftrightarrow4\left[\left(x-\frac{5}{8}\right)^2-\frac{89}{64}\right]=0\)
\(\Leftrightarrow4\left(x-\frac{5}{8}+\frac{\sqrt{89}}{8}\right)\left(x-\frac{5}{8}-\frac{\sqrt{89}}{8}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{5}{8}+\frac{\sqrt{89}}{8}=0\\x-\frac{5}{8}-\frac{\sqrt{89}}{8}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5-\sqrt{89}}{8}\\x=\frac{5+\sqrt{89}}{8}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{5-\sqrt{89}}{8};\frac{5+\sqrt{89}}{8}\right\}\)
Ta có x = 100
=> x + 1 = 101
Khi đó A = x15 - 101x14 + 101x13 - 101x12 + ... + 101x3 - 101x2 + 101x + 2020
= x15 - (x + 1)x14 + (x + 1)x13 - (x + 1)x12 + ... + (x + 1)x3 - (x + 1)x2 + (x + 1)x + 2020
= x15 - x15 - x14 + x14 + x13 - x13 - x12 + ... + x4 + x3 - x3 - x2 + x2 + x + 2020
= x + 2020
= 101 + 2020 (Vì x = 100)
= 2121
Vậy A = 2121 khi x = 100
A = x15 - 101x14 + 101x13 - ... - 101x2 + 101x + 2020 tại x = 100
x = 100 => 101 = x + 1
Thế vào A ta được
A = x15 - ( x + 1 )x14 + ( x + 1 )x13 - ... - ( x + 1 )x2 + ( x + 1 )x + 2020
= x15 - ( x15 + x14 ) + ( x14 + x13 ) - ... - ( x3 + x2 ) + ( x2 + x ) + 2020
= x15 - x15 - x14 + x14 + x13 - ... - x3 - x2 + x2 + x + 2020
= x + 2020
= 100 + 2020 = 2120
\(A=49x^2-28x+25\)
\(A=\left(7x\right)^2-2.7x.2+4-4+25\)
\(A=\left(7x-2\right)^2+21\)
Vì \(\left(7x-2\right)^2\ge0\) với mọi x
\(\Rightarrow\left(7x-2\right)^2+21\ge21\) với mọi x
\(\Rightarrow Amin=21\Leftrightarrow7x-2=0\)
\(\Rightarrow7x=2\)
\(\Rightarrow x=\dfrac{2}{7}\)
Vậy \(Amin=21\Leftrightarrow x=\dfrac{2}{7}\)
\(B=8x^2-28x-1\)
\(B=2\left(4x^2-14x-\dfrac{1}{2}\right)\)
\(B=2\left[\left(2x\right)^2-2.2x.\dfrac{7}{2}+\left(\dfrac{7}{2}\right)^2-\left(\dfrac{7}{2}\right)^2-\dfrac{1}{2}\right]\)
\(B=2\left[\left(2x\right)^2-2.2x.\dfrac{7}{2}+\left(\dfrac{7}{2}\right)^2-\dfrac{51}{4}\right]\)
\(B=2\left(2x-\dfrac{7}{2}\right)^2-\dfrac{51}{2}\)
Vì \(2\left(2x-\dfrac{7}{2}\right)^2\ge0\) với mọi x
\(\Rightarrow2\left(2x-\dfrac{7}{2}\right)^2-\dfrac{51}{2}\ge-\dfrac{51}{2}\)
\(\Rightarrow Bmin=-\dfrac{51}{2}\Leftrightarrow2x-\dfrac{7}{2}=0\)
\(\Rightarrow2x=\dfrac{7}{2}\)
\(\Rightarrow x=\dfrac{7}{4}\)
Vậy \(Bmin=-\dfrac{51}{2}\Leftrightarrow x=\dfrac{7}{4}\)
\(C=\left(2x^2+5\right)^2+10\)
Vì \(\left(2x^2+5\right)^2\ge0\) với mọi x
\(\Rightarrow\left(2x^2+5\right)^2+10\ge10\) với mọi x
\(\Rightarrow Cmin=10\Leftrightarrow2x^2+5=0\)
\(\Rightarrow2x^2=-5\)
\(\Rightarrow x^2=-\dfrac{5}{2}\)
\(\Rightarrow\) Không tồn tại x thỏa mãn
Vậy C không có giá trị nhỏ nhất
P/s: Câu c mình làm không có chắc nha, thấy nó sao sao ấy, không biết có sai đề không? 
\(D=3x^2-8x+7\)
\(D=3\left(x^2-\dfrac{8}{3}x+\dfrac{7}{3}\right)\)
\(D=3\left(x^2-2.x.\dfrac{4}{3}+\dfrac{16}{9}-\dfrac{16}{9}+\dfrac{7}{3}\right)\)
\(D=3\left(x^2-2.x.\dfrac{4}{3}+\dfrac{16}{9}+\dfrac{5}{9}\right)\)
\(D=3\left(x-\dfrac{4}{3}\right)^2+\dfrac{5}{3}\)
Vì \(3\left(x-\dfrac{4}{3}\right)^2\ge0\) với mọi x
\(\Rightarrow3\left(x-\dfrac{4}{3}\right)^2+\dfrac{5}{3}\ge\dfrac{5}{3}\)
\(\Rightarrow Dmin=\dfrac{5}{3}\Leftrightarrow x-\dfrac{4}{3}=0\)
\(\Rightarrow x=\dfrac{4}{3}\)
Vậy \(Dmin=\dfrac{5}{3}\Leftrightarrow x=\dfrac{4}{3}\)
\(E=x^4-2x^2+12\)
\(E=\left(x^2\right)^2-2x^2+1+11\)
\(E=\left(x^2-1\right)^2+11\)
Vì \(\left(x^2-1\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x^2-1\right)^2+11\ge11\) với mọi x
\(\Rightarrow Emin=11\Leftrightarrow x^2-1=0\)
\(\Rightarrow x^2=1\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy \(Emin=11\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
\(F=4x^2+15x+2\)
\(F=\left(2x\right)^2+2.2x.\dfrac{15}{4}+\left(\dfrac{15}{4}\right)^2-\left(\dfrac{15}{4}\right)^2+2\)
\(F=\left(2x+\dfrac{15}{4}\right)^2-\dfrac{225}{16}+\dfrac{32}{16}\)
\(F=\left(2x+\dfrac{15}{4}\right)^2-\dfrac{193}{16}\)
Vì \(\left(2x+\dfrac{15}{4}\right)^2\ge0\) với mọi x
\(\Rightarrow\left(2x+\dfrac{15}{4}\right)^2-\dfrac{193}{16}\ge-\dfrac{193}{16}\)
\(\Rightarrow Fmin=-\dfrac{193}{16}\Leftrightarrow2x+\dfrac{15}{4}=0\)
\(\Rightarrow2x=-\dfrac{15}{4}\)
\(\Rightarrow x=-\dfrac{15}{4}.\dfrac{1}{2}\)
\(\Rightarrow x=-\dfrac{15}{8}\)
Vậy \(Fmin=-\dfrac{193}{16}\Leftrightarrow x=-\dfrac{15}{8}\)
\(H=\left(x-1\right)\left(x+5\right)\left(x^2+4x+5\right)\)
\(H=\left(x^2+4x-5\right)\left(x^2+4x+5\right)\)
\(H=\left(x^2+4x\right)^2-5^2\)
\(H=\left(x^2+4x\right)^2-25\)
Vì \(\left(x^2+4x\right)^2\ge0\)
\(\Rightarrow\left(x^2+4x\right)^2-25\ge-25\) với mọi x
\(\Rightarrow Hmin=-25\Leftrightarrow x^2+4x=0\)
\(\Rightarrow x\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
Vậy \(Hmin=-25\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
\(I=\left(x^6+6\right)^2\)
Vì \(\left(x^6+6\right)^2\ge0\)
\(\Rightarrow Imin=0\Leftrightarrow x^6+6=0\)
\(\Rightarrow\left(x^3\right)^2=-6\)
\(\Rightarrow\) Không tồn tại x
Vậy I không có giá trị nhỏ nhất
\(A=49x^2-28x+25=\left(49x^2-28x+1\right)+24=\left(7x-1\right)^2+24\ge24\)
Vậy GTNN của A là 24 khi x = \(\dfrac{1}{7}\)
\(B=8x^2-28x-1=8\left(x^2-\dfrac{7}{2}x+\dfrac{49}{16}\right)-\dfrac{51}{2}=8\left(x-\dfrac{7}{4}\right)^2-\dfrac{51}{2}\ge-\dfrac{51}{2}\)
Vậy GTNN của B là \(-\dfrac{51}{2}\) khi x = \(\dfrac{7}{4}\)
\(C=\left(2x^2+5\right)^2+10=4x^4+20x^2+35\ge35\)
Vậy GTNN của C là 35 khi x = 0
\(D=3x^2-8x+7=3\left(x^2-\dfrac{8}{3}x+\dfrac{16}{9}\right)+\dfrac{5}{3}=3\left(x-\dfrac{4}{3}\right)^2+\dfrac{5}{3}\ge\dfrac{5}{3}\)
Vậy GTNN của D là \(\dfrac{5}{3}\) khi x = \(\dfrac{4}{3}\)
\(E=x^4-2x^2+12=\left(x^4-2x^2+1\right)+11=\left(x^2-1\right)^2+11\ge11\)
Vậy GTNN của E là 11 khi x = 1 hoặc x = -1
\(F=4x^2+15x+2=\left(4x^2+15x+\dfrac{225}{16}\right)-\dfrac{193}{16}=\left(2x+\dfrac{15}{4}\right)^2-\dfrac{193}{16}\ge-\dfrac{193}{16}\)
Vậy GTNN của F là \(-\dfrac{193}{16}\) khi x = \(-\dfrac{15}{8}\)
\(G=8\left(a+2\right)^3-\left(2a+1\right)^3\)
\(G=36a^2+90a+63\)
\(G=9\left(4a^2+10a+7\right)\)
\(G=9\left(4a^2+10a+\dfrac{25}{4}\right)+\dfrac{27}{4}\)
\(G=9\left(2a+\dfrac{5}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}\)
Vậy GTNN của G là \(\dfrac{27}{4}\) khi x = \(-\dfrac{5}{4}\)
\(H=\left(x-1\right)\left(x+5\right)\left(x^2+4x+5\right)\)
\(H=x^4+8x^3+16x^2-25\)
\(H=\left(x^2+4x\right)^2-25\ge-25\)
Vậy GTNN của H là -25 khi x = -4 hoặc x = 0
\(I=\left(x^6+6\right)^2=x^{12}+12x^6+36\ge36\)
Vậy GTNN của I là 36 khi x = 0
Làm 2 câu các câu còn lại tương tự!
a, \(E=-x^2+4x-5=-\left(x^2-4x+5\right)\)
\(=-\left(x^2-2x-2x+4+1\right)=-\left[\left(x-2\right)^2+1\right]\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(x-2\right)^2+1\ge1\Rightarrow-\left[\left(x-2\right)^2+1\right]\le-1\)
Hay \(E\le-1\) với mọi giá trị của \(x\in R\).
Để \(E=-1\) thì \(-\left[\left(x-2\right)^2+1\right]=-1\)
\(\Rightarrow\left(x-2\right)^2=0\Rightarrow x=2\)
Vậy.............
b, \(F=-2x^2+2x-1=-\left(2x^2-2x+1\right)\)
\(=-\left(2x^2-x-x+\dfrac{1}{2}-\dfrac{3}{2}\right)\)
\(=-\left[\left(2x-1\right)^2-\dfrac{3}{2}\right]\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(2x-1\right)^2-\dfrac{3}{2}\ge-\dfrac{3}{2}\Rightarrow-\left[\left(2x-1\right)^2-\dfrac{3}{2}\right]\le\dfrac{3}{2}\)
Hay \(F\le\dfrac{3}{2}\) với mọi giá trị của \(x\in R\).
Để \(F=\dfrac{3}{2}\) thì \(-\left[\left(2x-1\right)^2-\dfrac{3}{2}\right]=\dfrac{3}{2}\)
\(\Rightarrow\left(2x-1\right)^2=0\Rightarrow x=\dfrac{1}{2}\)
Vậy.............
7, \(G=-4x^2+12x-7\)
\(=-4\left(x^2-3x+\dfrac{7}{4}\right)\)
\(=-4\left(x^2-\dfrac{3}{2}.x.2+\dfrac{9}{4}-\dfrac{2}{4}\right)\)
\(=-4\left(x-\dfrac{3}{2}\right)^2+2\le2\)
Dấu " = " khi \(-4\left(x-\dfrac{3}{2}\right)^2=0\Leftrightarrow x=\dfrac{3}{2}\)
Vậy \(MAX_G=2\) khi \(x=\dfrac{3}{2}\)
8, \(H=-2x^2+4x-15\)
\(=-2\left(x^2-2x+\dfrac{15}{2}\right)\)
\(=-2\left(x^2-2x+1+\dfrac{13}{2}\right)\)
\(=-2\left(x-1\right)^2-13\le-13\)
Dấu " = " khi \(-2\left(x-1\right)^2=0\Leftrightarrow x=1\)
Vậy \(MAX_H=-13\) khi x = 1
9, \(K=-x^4+2x^2-2\)
\(=-\left(x^2-2x^2+1+1\right)\)
\(=-\left(x^2-1\right)^2-1\le-1\)
Dấu " = " khi \(-\left(x^2-1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy \(MAX_K=-1\) khi \(x=\pm1\)
10, \(J=-3x^2+15x-9\)
\(=-3\left(x^2-\dfrac{5}{2}.x.2+\dfrac{10}{4}+\dfrac{2}{4}\right)\)
\(=-3\left(x-\dfrac{5}{2}\right)^2-\dfrac{3}{2}\le\dfrac{-3}{2}\)
Dấu " = " khi \(-3\left(x-\dfrac{5}{2}\right)^2=0\Leftrightarrow x=\dfrac{5}{2}\)
Vậy \(MAX_J=\dfrac{-3}{2}\) khi \(x=\dfrac{5}{2}\)
\(f\left(100\right)\Leftrightarrow x=100\)
\(\Rightarrow x+1=101\left(1\right)\)
Thay (1) vào ta được
\(f\left(100\right)=x^8-\left(x+1\right)x^7+\left(x+1\right)x^6-\left(x+1\right)x^5+...+\left(x+1\right)x^2-\left(x+1\right)x+25\)
\(f\left(100\right)=x^8-x^8-x^7+x^7+x^6-x^6-x^5+...+x^2-x^2-x+25\)
\(f\left(100\right)=-x+25\)
\(f\left(100\right)=-100+25\)
\(f\left(100\right)=-75\)