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đk : x khác 2; x khác 3; x khác 1
\(a.A=\left(\frac{x^2}{x^2-5x+6}+\frac{x^2}{x^2-3x+2}\right)\cdot\frac{x^2-4x+3}{x^4+x^2+1}\)
\(A=\left(\frac{x^2}{\left(x-2\right)\left(x-3\right)}+\frac{x^2}{\left(x-1\right)\left(x-2\right)}\right)\cdot\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
\(A=\left(\frac{x^2\left(x-1\right)+x^2\left(x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\right)\cdot\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
\(A=\frac{x^2\left(x-1+x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\cdot\frac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
\(A=\frac{x^2\left(2x-4\right)}{\left(x-2\right)\left(x^4+x^2+1\right)}=\frac{2x^2}{x^4+x^2+1}\)
\(b.\frac{1}{A}=\frac{x^4+x^2+1}{2x^2}=\frac{x^2}{2}+\frac{1}{2}+\frac{1}{2x^2}\) (x khác 0)
\(\frac{1}{A}=\frac{2x^2}{4}+\frac{1}{2}+\frac{1}{2x^2}\)
có 2x^2/4 và 1/2x^2 > 0 áp dụng bđt cô si ta có
\(\frac{2x^2}{4}+\frac{1}{2x^2}\ge2\sqrt{\frac{2x^2}{4}\cdot\frac{1}{2x^2}}=1\)
\(\Rightarrow\frac{1}{A}\ge\frac{3}{2}\)
\(\Rightarrow A\le\frac{2}{3}\)
DẤU = xảy ra khi 2x^2/4 = 1/2x^2 => 4x^4 = 4
=> x^4 = 1
=> x = 1 (loại) hoặc x = -1 (thỏa mãn)
vậy max a = 2/3 khi x = -1
a, \(A=\left(\frac{4}{2x+1}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\left(\frac{4\left(x^2+1\right)}{\left(2x+1\right)\left(x^2+1\right)}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\left(\frac{4x^2+4+4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\frac{\left(2x+1\right)^2}{\left(x^2+1\right)\left(2x+1\right)}\frac{x^2+1}{x^2+2}=\frac{2x+1}{x^2+2}\)
\(A=x^2-4x-x\left(x-4\right)-15\)
\(=x^2-4x-x^2+4x-15=-15\) => đpcm
\(B=5x\left(x^2-x\right)-x^2\left(5x-5\right)-13\)
\(=5x^3-5x^2-5x^3+5x^2-13=-13\) => đpcm
\(C=-3x\left(x-5\right)+3\left(x^2-4x\right)-3x+7\)
\(=-3x^2+15x+3x^2-12x-3x+7=7\) => đpcm
\(D=7\left(x^2-5x+3\right)-x\left(7x-35\right)-14\)
\(=7x^2-35x+21-7x^2+35x-14=7\) => đpcm
\(E=4x\left(x^2-7+2\right)-4\left(x^3-7x+2x-5\right)\)
\(=4x^3-20x-4x^3+20x+20=20\) => đpcm
\(H=x\left(5x-3\right)-x^2\left(x-1\right)+x\left(x^2-6x\right)-10+3x\)
\(=5x^2-3x-x^3+x^2+x^3-6x^2-10x+3x=-10\) => đpcm
\(A=x^2-3x+5\)
\(=x^2-3x+\frac{9}{4}+\frac{11}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)
\(\left(x-\frac{3}{2}\right)^2\ge0\Rightarrow A\ge\frac{11}{4}\)
Dấu "=" xảy ra khi \(x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)
Vậy Min A = \(\frac{11}{4}\Leftrightarrow x=\frac{3}{2}\)
a) \(A=x^2-3x+5\)
\("="\Leftrightarrow x=\frac{11}{4}\Rightarrow x=\frac{3}{2};\frac{11}{4}\)
b) \(B=\left(2x-1\right)^2+\left(x+2\right)^2\)
\("="\Leftrightarrow x=5\Rightarrow x=0;5\)
c) \(C=4x-x^2+3\)
\("="\Leftrightarrow x=7\Rightarrow x=2;7\)
d) \(D=x^4+x^2+2\)
\("="\Leftrightarrow x=2\Rightarrow x=0;2\)
Mình nghĩ bạn viết hơi sai đề bài.
\(x^2+xz-y^2-yz=\left(x^2-y^2\right)+xz-yz=\left(x-y\right)\left(x+y\right)+z\left(x-y\right)=\left(x-y\right)\left(x+y+z\right)\)
Tương tự: \(y^2+xy-z^2-xz=\left(y-z\right)\left(x+y+z\right)\)
\(z^2+yz-x^2-xy=\left(x+y+z\right)\left(z-x\right)\)
Khi đó:
\(P=\frac{1}{\left(y-z\right)\left(x-y\right)\left(x+y+z\right)}+\frac{1}{\left(z-x\right)\left(y-z\right)\left(x+y+z\right)}+\frac{1}{\left(x-y\right)\left(x+y+z\right)\left(z-x\right)}\)
\(=\frac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x+y+z\right)}=0\)
\(a,\left(x+2\right)^2-4x\left(x-1\right)=-3x\left(x-6\right)\)
\(\Leftrightarrow x^2+4x+4-4x^2+4x=-3x^2+18x\)
\(\Leftrightarrow x^2-4x^2+3x^2+4x+4x-18x=-4\)
\(\Leftrightarrow-10x=-4\)
\(\Leftrightarrow x=\frac{2}{5}\)
Câu b làm tương tự nhé
Học tốt ##
\(b,\left(x+3\right)^3-\left(x-2\right)^3=\left(5x-1\right)\left(3x+2\right)\)
\(\Rightarrow x^3+6x^2+9x+3x^2+18x+27-x^3+4x^2-4x+2x^2-8x+8=15x^2+10x-3x-2\)
\(\Rightarrow15x^2+15x+35=15x^2+10x-3x-2\)
\(\Rightarrow15x^2+15x+35=15x^2+7x-2\)
\(\Rightarrow15x+35=7x-2\)
\(\Rightarrow15x-7x=-2-35\)
\(\Rightarrow8x=-37\)
\(\Rightarrow x=\frac{-37}{8}\)
a/ \(A=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left[\left(x+1\right)\left(x-6\right)\right].\left[\left(x-2\right)\left(x-3\right)\right]\)
\(=\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\ge-36\)
Suy ra Min A = -36 <=> \(x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=5\end{array}\right.\)
b/ \(B=19-6x-9x^2=-9\left(x-\frac{1}{3}\right)^2+20\le20\)
Suy ra Min B = 20 <=> x = 1/3
a) \(A=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\)
\(=\left[\left(x+1\right)\left(x-6\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]\)
\(\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\)
Vì \(\left(x^2-5x\right)^2\ge0\)
=> \(\left(x^2-5x\right)^2-36\ge-36\)
Vậy GTNN của A là -36 khi \(x^2-5x=0\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=5\end{array}\right.\)
b) \(B=19-6x-9x^2=-\left(9x^2+6x+1\right)+20=-\left(3x+1\right)^2+20\)
Vì \(-\left(3x+1\right)^2\le0\)
=> \(-\left(3x+1\right)+20\le20\)
Vậy GTLN của B là 20 khi \(x=-\frac{1}{3}\)
Bài 1:
a)(4x-3)(3x+2)-(6x+1)(2x-5)+1
=12x2-x-6-12x2+28x+5+1
=27x
b)(3x+4)2+(4x-1)2+(2+5x)(2-5x)
=9x2+24x+16+16x2-8x+1+4-25x2
=16x+21
c)(2x+1)(4x2-2x+1)+(2-3x)(4+6x+9x2)-9
=8x3+1+8-27x3-9
=-19x3
Bài 2:
a)3x(x-4)-x(5+3x)=-34
=>3x2-12x-3x2-5x=-34
=>-17x=-34
=>x=2
Vậy x=2
b)(3x+1)2+(5x-2)2=34(x+2)(x-2)
=>9x2+6x+1+25x2-20x+4=34(x2-4)
=>34x2-14x+5-34x2+136=0
=>-14x+141=0
=>-14x=-141
=>x=\(\frac{141}{14}\)
Vậy x=\(\frac{141}{14}\)
c)x3+3x2+3x+28=0
=>x3-x2+7x+4x2-4x+28=0
=>x(x2-x+7)+4(x2-x+7)=0
=>(x+4)(x2-x+7)=0
\(\Rightarrow\left[\begin{array}{nghiempt}x+4=0\\x^2-x+7=0\left(2\right)\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-4\\\left(2\right)\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{27}{4}>0\end{array}\right.\)
=>(2) vô nghiệm
Vậy x=-4
a) \(A\left(x\right)=-5x^2-4x+1\)
\(\Leftrightarrow A\left(x\right)=-\left(5x^2+4x-1\right)\)
\(\Leftrightarrow A\left(x\right)=-\left[\left(\sqrt{5}x\right)^2+2.\sqrt{5}x.\dfrac{2\sqrt{5}}{5}+\dfrac{4}{5}-\dfrac{9}{5}\right]\)
\(\Leftrightarrow A\left(x\right)=-\left(\sqrt{5}x+\dfrac{4}{5}\right)^2+\dfrac{9}{5}\le\dfrac{9}{5}\)
Dấu bằng xảy ra
\(\Leftrightarrow\sqrt{5}x+\dfrac{4}{5}=0\Leftrightarrow x==-\dfrac{4\sqrt{5}}{25}\)
b) \(B\left(x\right)=-3x^2+x+1\)
\(\Leftrightarrow B\left(x\right)=-\left[\left(\sqrt{3}x\right)^2-2.\sqrt{3}x.\dfrac{2\sqrt{3}}{3}+\left(\dfrac{2\sqrt{3}}{3}\right)^2-\dfrac{1}{3}\right]\)
\(\Leftrightarrow B\left(x\right)=-\left(\sqrt{3}x-\dfrac{2\sqrt{3}}{3}\right)^2+\dfrac{1}{3}\le\dfrac{1}{3}\)
Dấu bằng xảy ra
\(\Leftrightarrow\sqrt{3}x-\dfrac{2\sqrt{3}}{3}=0\Leftrightarrow x=\dfrac{2}{3}\)
A = -5x2 -4x + 1
A = - ( 5x2 + 2.\(\sqrt{5}\).\(\dfrac{2}{\sqrt{5}}\)x +\(\dfrac{4}{5}\) ) +\(\dfrac{29}{25}\)
A = -( \(\sqrt{5}\) x+ \(\dfrac{2}{\sqrt{5}}\))2 + \(\dfrac{29}{25}\)
-( \(\sqrt{5}\) x+ \(\dfrac{2}{\sqrt{5}}\))2 ≤ 0 ⇔ A(max) =\(\dfrac{29}{25}\) ⇔ x = -2/5
B = -3x2 + x + 1
B = -(3x2 - 2.\(\sqrt{3}\).\(\dfrac{1}{2\sqrt{3}}\).x + \(\dfrac{1}{12}\)) + \(\dfrac{13}{12}\)
B = -(\(\sqrt{3}\)x - \(\dfrac{1}{2\sqrt{3}}\))2 + \(\dfrac{13}{12}\)
vì - (\(\sqrt{3}\)x - \(\dfrac{1}{2\sqrt{3}}\))2 ≤ 0 ⇔ B(max) =\(\dfrac{13}{12}\) ⇔ x = \(\dfrac{1}{6}\)