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\(A=-x^2+6x-6=-\left(x^2-6x+9\right)+3=-\left(x-3\right)^2+3\le3 \)
Vậy GTLN của A là 3 khi x = 3
\(B=-2x^22+5x-10=-\left(4x^2-5x+\frac{25}{16}\right)-\frac{135}{16}=-\left(2x-\frac{5}{4}\right)-\frac{135}{16}\le-\frac{135}{16}\)
Vậy GTLN của B là \(-\frac{135}{16}\)khi x = \(\frac{5}{8}\)
\(C=-5x^2+x+15=-5\left(x^2-\frac{1}{5}x+\frac{1}{100}\right)+\frac{301}{20}=-5\left(x-\frac{1}{10}\right)^2+\frac{301}{20}\le\frac{301}{20}\)
Vậy GTLN của C là \(\frac{301}{20}\)khi x = \(\frac{1}{10}\)
Bài 1:
\(P=3x^2+x-1\)
\(=3\left(x^2+\frac{1}{3}x-\frac{1}{3}\right)\)
\(=3\left(x^2+2x.\frac{1}{6}+\frac{1}{36}-\frac{13}{36}\right)\)
\(=3\left(x+\frac{1}{6}\right)^2-\frac{13}{12}\ge\frac{-13}{12}\)\(\forall x\)
Dấu '' = '' xảy ra khi: \(\left(x+\frac{1}{6}\right)^2=0\Rightarrow x=\frac{-1}{6}\)
Vậy \(MinP=\frac{-13}{12}\) khi \(x=\frac{-1}{6}\)
Bài 2:
a) Không có điều kiện
b) Nghiệm vô tỉ
Bạn xem lại đề hai phần này nhé.
c) \(\left(x-2\right)^3-x^3+6x^2=14\)
\(\Rightarrow x^3-6x^2+12x-8-x^3+6x^2-14=0\)
\(\Rightarrow\left(x^3-x^3\right)+\left(-6x^2+6x^2\right)+12x+\left(-8-14\right)=0\)
\(\Rightarrow12x-22=0\)
\(\Rightarrow x=\frac{11}{6}\)
d) \(8x^2+30x+7=0\)
\(\Rightarrow8x^2+28x+2x+7=0\)
\(\Rightarrow\left(8x^2+28x\right)+\left(2x+7\right)=0\)
\(\Rightarrow4x\left(2x+7\right)+\left(2x+7\right)=0\)
\(\Rightarrow\left(4x+1\right)\left(2x+7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4x+1=0\\2x+7=0\end{cases}}\Rightarrow\orbr{\begin{cases}4x=-1\\2x=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{4}\\x=\frac{-7}{2}\end{cases}}\)
Tìm GTNN
a/ \(A=4x^2+7x+13=\left(4x^2+7x+\frac{49}{16}\right)+\frac{159}{16}=\left(2x+\frac{7}{4}\right)^2+\frac{159}{16}\ge\frac{159}{16}\)
b/ \(B=5-8x+x^2=\left(x^2-8x+16\right)-11=\left(x-4\right)^2-11\ge-11\)
c/ \(C=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\)
\(A=\left(x-1\right)^2+\left(x-2\right)^2+5.\)
\(A=\left(x^2-2.x.1+1^2\right)+\left(x^2-2.x.2+2^2\right)+5.\)
\(A=\left(x^2-2x+1\right)+\left(x^2-4x+4\right)+5.\)
( suy nghĩ tiếp nha)
Hok tốt
\(A=\left(x-1\right)\left(x-5\right)+18\)
\(=x^2-6x+5+18\)
\(=x^2-6x+9+14\)
\(=\left(x-3\right)^2+14\)
\(\Rightarrow A_{min}=14\Leftrightarrow\left(x-3\right)^2=0\)
\(\Rightarrow x-3=0\Leftrightarrow x=3\)
Ta có:
A = (x - 1)(x - 5) + 18 = x2 - 5x - x + 5 + 18 = x2 - 6x + 23 = (x2 - 6x + 9) + 14 = (x - 3)2 + 14
Ta luôn có: (x - 3)2 \(\ge\)0 \(\forall\)x => (x - 3)2 + 14 \(\ge\)14 \(\forall\)x
hay A \(\ge\)14 \(\forall\)x
Dấu "=" xảy ra khi : (x - 3)2 = 0 <=> x - 3 = 0 <=> x = 3
Vậy Amin = 14 tại x = 3
1. Câu hỏi của Quỳnh Như - Toán lớp 8 - Học toán với OnlineMath
Em tham khảo câu 1 tại link này.
\(A=x^2-6x+10=x^2-2.3x+3^2+1=\left(x-3\right)^2+1\)
Ta có: \(\left(x-3\right)^2\ge0\) nên \(\left(x-3\right)^2+1\ge1\)
Vậy \(A_{min}=1\)(Dấu "="\(\Leftrightarrow x=3\))
a) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)
\(\Leftrightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3+3x^2\right)=2\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3-3x^2=2\)
\(\Leftrightarrow3x+1=2\)
\(\Leftrightarrow3x=1\)
\(\Leftrightarrow x=\frac{1}{3}\)
\(A=-x^2-5y^2+2xy-4x+20y+13\)
\(=-x^2+2xy-y^2-4y^2-4x+4y+16y+13\)
\(=-\left(x^2-2xy+y^2\right)-\left(4y^2-16y+16\right)-\left(4x-4y\right)+29\)
\(=-\left(x-y\right)^2-4\left(y-2\right)^2-4\left(x-y\right)-4+25\)
\(=-\left[\left(x-y\right)^2+4\left(x-y\right)+4\right]-4\left(y-2\right)^2+25\)
\(=-\left(x-y+2\right)^2-4\left(y-2\right)^2+25\)
\(A_{max}=25\Leftrightarrow\hept{\begin{cases}\left(x-y+2\right)^2=0\\\left(y-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x-y+2=0\\y=2\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}x=0\\y=2\end{cases}}\)
\(B=-7x^2-y^2+4xy+16x-2y+17.\)
\(=-4x^2+4xy-y^2-3x^2+12x-12+4x-2y+29\)
\(=-\left(2x-y\right)^2-3\left(x-2\right)^2+2\left(2x-y\right)^2-1+30\)
\(=-\left[\left(2x-y\right)^2-2\left(2x-y\right)^2+1\right]-3\left(x-2\right)^2+30\)
\(=-\left(2x-y-1\right)^2-3\left(x-2\right)^2+30\)
\(\Rightarrow B_{max}=30\Leftrightarrow\hept{\begin{cases}\left(2x-y-1\right)^2=0\\\left(x-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}2x-y-1=0\\x=2\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}\)
\(a.A=x^2+5x+7\\ =\left[x^2+2\cdot x\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]+\dfrac{3}{4}\\ =\left(x+\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu "=" xảy ra: `x+5/2=0<=>x=-5/2`
\(b.B=6x-x^2-5\\ =-\left(x^2-6x+9\right)+4\\ =-\left(x-3\right)^2+4\le4\forall x\)
Dấu "=" xảy ra: `x-3=0<=>x=3`