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MN
1
7 tháng 10 2020
a) \(A=x^2+6x+1=\left(x^2+2\cdot x\cdot3+3^2\right)-8\)
\(=\left(x+3\right)^2-8\)
Vì \(\left(x+3\right)^2\ge0\forall x\)
=> \(\left(x+3\right)^2-8\ge-8\forall x\)
Dấu " = " xảy ra khi và chỉ khi (x + 3)2 = 0 => x = -3
Vậy Amin = -8 khi x = -3
b) \(2x^2+10x-5=2\left(x^2+5x-\frac{5}{2}\right)\)
\(=2\left[x^2+2\cdot x\cdot\frac{5}{2}+\left(\frac{5}{2}\right)^2\right]-\frac{35}{2}\)
\(=2\left(x+\frac{5}{2}\right)^2-\frac{35}{2}\)
Vì (x + 5/2)2 \(\ge0\forall x\)
=> \(2\left(x+\frac{5}{2}\right)^2-\frac{35}{2}\ge-\frac{35}{2}\forall x\)
Dấu " = " xảy ra khi và chỉ khi (x + 5/2)2 = 0 => x = -5/2
Vậy Bmin = -35/2 khi x = -5/2
c) \(x^2-5x=\left[x^2-2\cdot x\cdot\frac{5}{2}+\left(\frac{5}{2}\right)^2\right]-\frac{25}{4}\)
\(=\left(x-\frac{5}{2}\right)^2-\frac{25}{4}\)
Vì (x - 5/2)2 \(\ge\)0 với mọi x
=> \(\left(x-\frac{5}{2}\right)^2-\frac{25}{4}\ge-\frac{25}{4}\)
Dấu " = " xảy ra khi và chỉ khi (x - 5/2)2 = 0 => x = 5/2
Vậy Cmin = -25/4 khi x = 5/2
16 tháng 11 2019
Ta có:
\(\frac{2}{x^2+2x}+\frac{2}{x^2+6x+8}+\frac{2}{x^2+10x+24}+\frac{1}{x+6}\)
= \(\frac{2}{x\left(x+2\right)}+\frac{2}{x^2+4x+2x+8}+\frac{2}{x^2+4x+6x+24}+\frac{1}{x+6}\)
= \(\frac{2}{x\left(x+2\right)}+\frac{2}{x\left(x+4\right)+2\left(x+4\right)}+\frac{2}{x\left(x+4\right)+4\left(x+6\right)}+\frac{1}{x+6}\)
= \(\frac{2}{x\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{2}{\left(x+4\right)\left(x+6\right)}+\frac{1}{x+6}\)
= \(\frac{1}{x}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+6}+\frac{1}{x+6}\)
= \(\frac{1}{x}\)
KN
18 tháng 8 2020
1. \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
\(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
\(\Leftrightarrow35x-5+60x=96-6x\)
\(\Leftrightarrow95x-5=96-6x\)
\(\Leftrightarrow95x+6x=96+5\)
\(\Leftrightarrow101x=101\)
\(\Leftrightarrow x=1\)
2. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(\Leftrightarrow3\left(10x+3\right)=36+4\left(6+8x\right)\)
\(\Leftrightarrow30x+9=36+24+32x\)
\(\Leftrightarrow30x+9=32x+60\)
\(\Leftrightarrow30x-32x=60-9\)
\(\Leftrightarrow-2x=51\)
\(\Leftrightarrow x=-\frac{51}{2}\)
3. \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow8x-3-2\left(3x-2\right)=2\left(2x-1\right)+x+3\)
\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)
\(\Leftrightarrow2x+1=5x+1\)
\(\Leftrightarrow2x=5x\)
\(\Leftrightarrow x=0\)
19 tháng 8 2020
4) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
=> \(\frac{9-3x}{8}+\frac{10-2x}{3}=\frac{1-x}{2}-\frac{2}{1}\)
=> \(\frac{3\left(9-3x\right)}{24}+\frac{8\left(10-2x\right)}{24}=\frac{12\left(1-x\right)}{24}-\frac{48}{24}\)
=> \(\frac{27-9x}{24}+\frac{80-16x}{24}=\frac{12-12x}{24}-\frac{48}{24}\)
=> \(\frac{27-9x+80-16x}{24}=\frac{12-12x-48}{24}\)
=> 27 - 9x + 80 - 16x = 12 - 12x - 48
=> 27 - 9x + 80 - 16x - 12 + 12x + 48 = 0
=> (27 + 80 - 12 + 48) + (-9x - 16x + 12x) = 0
=> 143 - 13x = 0
=> 13x = 143
=> x = 11
5) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
=> \(\frac{2x-6}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
=> \(\frac{3\left(2x-6\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)
=> \(\frac{6x-18}{21}+\frac{7x-35}{21}-\frac{13x+4}{21}=0\)
=> \(\frac{6x-18+7x-35-13x-4}{21}=0\)
=> 6x - 18 + 7x - 35 - 13x - 4 = 0
=> (6x + 7x - 13x) + (-18 - 35 - 4) = 0
=> -57 = 0(vô nghiệm)
6) \(\frac{6x+5}{2}-\left(2x+\frac{2x+1}{2}\right)=\frac{10x+3}{4}\)
=> \(\frac{6x+5}{2}-\frac{10x+3}{4}=2x+\frac{2x+1}{2}\)
=> \(\frac{2\left(6x+5\right)}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{2\left(2x+1\right)}{4}\)
=> \(\frac{12x+10}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{4x+2}{4}\)
=> \(\frac{12x+10-\left(10x+3\right)}{4}=\frac{8x+4x+2}{4}\)
=> \(\frac{12x+10-10x-3}{4}=\frac{12x+2}{4}\)
=> \(12x+10-10x-3=12x+2\)
=> \(2x+10-3=12x+2\)
=> 2x + 10 - 3 - 12x - 2 = 0
=> (2x - 12x) + (10 - 3 - 2) = 0
=> -10x + 5 = 0
=> -10x = -5
=> x = 1/2
7) \(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)
=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{x+7}{15}=0\)
=> \(\frac{6x-3}{15}-\frac{5x-10}{15}-\frac{x+7}{15}=0\)
=> \(\frac{6x-3-\left(5x-10\right)-\left(x+7\right)}{15}=0\)
=> 6x - 3 - 5x + 10 - x - 7 = 0
=> (6x - 5x - x) + (-3 + 10 - 7) = 0
=> 0x + 0 = 0
=> 0x = 0
=> x tùy ý
Bài 8 tự làm nhé
1 tháng 7 2017
a ) MTC : \(2x\left(x+3\right)\left(x-3\right)\)
\(\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{3-2x}{x^2-9}=\frac{3-2x}{\left(x-3\right)\left(x+3\right)}=\frac{2x\left(3-2x\right)}{2x\left(x+3\right)\left(x-3\right)}\)
b ) MTC : \(2\left(-x\right)\left(x-1\right)^2\)
\(\frac{2x-1}{x-x^2}=\frac{2x-1}{-x\left(x-1\right)}=\frac{2\left(2x-1\right)\left(x-1\right)}{2\left(-x\right)\left(x-1\right)^2}\)
\(\frac{x+1}{2-4x+2x^2}=\frac{x+1}{2\left(x^2-2x+1\right)}=\frac{-x\left(x+1\right)}{2\left(-x\right)\left(x-1\right)^2}\)
6 tháng 5 2017
a) \(B=\frac{3x^2+6x+10}{x^2+2x+5}\)
\(\Leftrightarrow B=3-\frac{5}{x^2+2x+5}\)
\(\Leftrightarrow B=3-\frac{5}{5\left(\frac{x^2}{5}+\frac{2x}{5}+\frac{5}{5}\right)}\Leftrightarrow B=3-\frac{1}{\frac{\left(x^2+2x+1\right)}{5}+\frac{4}{5}}\)( cho \(\left(x+1\right)^2=0\))
\(\Leftrightarrow maxB=3-\frac{1}{\frac{4}{5}}=\frac{7}{4}\) KHI X= -1
c) \(D=x^2-2x+y^2+4y+7\)
\(\Leftrightarrow D=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+2\)
\(\Leftrightarrow D=\left(x-1\right)^2+\left(y+2\right)^2+2\)
\(\Leftrightarrow minD=2\)KHI X= 1 và Y= -2
e) Câu này đề có vẻ sai bạn kiểm tra lại giúp mk ! mk làm theo đề đúng nka !
\(E=\frac{x^2-4x+1}{x^2}\)
\(\Leftrightarrow E=\frac{x^2\left(1-\frac{4}{x}+\frac{1}{x^2}\right)}{x^2}=1-\frac{4}{x}+\frac{1}{x^2}\)
ĐẶT \(y=\frac{1}{x}\)\(\Leftrightarrow minE=-3\)KHI X = 1/2
Hai câu còn lại tối mk giải tiếp mk bận đi học rùi bạn thông cảm
HN
2
2 tháng 5 2017
đáng lẽ phải là x^2+2x+3 chứ bạn
y-1=(3x^2+10x+11)/(x^2+2x+3)-1
y-1=(3x^2+10+11-x^2-2x-3)/(x^2+2x+3)
y-1=(2x^2+8x+8)/(x^2+2x+3)
y-1=2(x+2)^2/(x^2+2x+3)>=0
y>=1
=>Min y=1 khi x+2=0 hay x=-2
y-4=(3x^2+10x+11)/(x^2+2x+3)-4
y-4=(3x^2+10x+11-4x^2-8x-12)/(x^2+2x+3)
y-4=(-x^2+2x-1)/(x^2+2x+3)
y-4=-(x-1)^2/(x^2+2x+3)<=0
y<=4
=>Max y=4 khi x-1=0 hay x=1
\(Q=\frac{10x^2+6x+3}{x^2+2}\)
\(Q=\frac{x^2+2+9x^2+6x+1}{x^2+2}\)
\(Q=\frac{x^2+2}{x^2+2}+\frac{9x^2+6x+1}{x^2+2}\)
\(Q=1+\frac{\left(3x+1\right)^2}{x^2+2}\)
Vì \(\left(3x+1\right)^2\ge0\forall x\)
\(\Rightarrow Q\ge1+0=1\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow3x+1=0\Leftrightarrow x=\frac{-1}{3}\)
Vậy \(Q_{min}=1\Leftrightarrow x=\frac{-1}{3}\)