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a) \(A=25x^2+3y^2-10x+11\)
\(A=\left(5x-1\right)^2+3y^2+11\ge11\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{5}\\y=0\end{matrix}\right.\)
b) \(B=\left(x-3\right)^2+\left(x-11\right)^2\)
\(B=2\left(x^2-14x+65\right)\)
\(B=2\left[\left(x-7\right)^2+16\right]\)
\(B=2\left(x-7\right)^2+32\ge32\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=7\)
c) \(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\)
\(C=\left(x^2-5x-6\right)\left(x^2-5x+6\right)\)
Đặt \(x^2-5x-6=a\)
\(C=a\left(a+12\right)\)
\(C=a^2+12a+36-36\)
\(C=\left(a+6\right)^2-36\ge-36\)
Dấu "=" xảy ra \(\Leftrightarrow a=-6\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
\(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\\ C=\left(x+1\right)\left(x-6\right)\left(x-2\right)\left(x-3\right)\\ C=\left(x^2-5x-6\right)\left(x^2-5x+6\right)\\ C=\left(x^2-5x\right)^2-6^2\\ C=\left(x^2-5x\right)^2-36\)
Ta có:
\(\left(x^2-5x\right)^2\ge0\\ \Rightarrow C=\left(x^2-5x\right)^2-36\ge-36\)
Dấu "=" xảy ra khi và chỉ khi:
(x2 - 5x)2 = 0 => x2 - 5x = 0 => x(x - 5) = 0
=> x = 5 hoặc x = 0
Vậy MinC = -36 <=> x = 5; x = 0
1)
a) \(\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2+2\left(ab^2c+a^2bc+abc^2\right)\)\(=a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=a^2b^2+b^2c^2+c^2a^2\)(vì a+b+c=0)
b) \(a+b+c=0\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right]\)
\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)=2\left(ab+bc+ca\right)^2\left(theoa\right)\)
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
a) (x-1)(5x+3)=(3x-8)(x-1)
= (x-1)(5x+3)-(3x-8)(x-1)=0
=(x-1)[(5x+3)-(3x-8)]=0
=(x-1)(5x+3-3x+8)=0
=(x-1)(2x+11)=0
\(\Leftrightarrow\) x-1=0 hoặc 2x+11=0
\(\Leftrightarrow\) x=1 hoặc x=\(\dfrac{-11}{2}\)
Vậy S={1;\(\dfrac{-11}{2}\)}
b) 3x(25x+15)-35(5x+3)=0
=3x.5(5x+3)-35(5x+3)=0
=15x(5x+3)-35(5x+3)=0
=(5x+3)(15x-35)=0
\(\Leftrightarrow\) 5x+3=0 hoặc 15x-35=0
\(\Leftrightarrow\) x=\(\dfrac{-3}{5}\) hoặc x=\(\dfrac{7}{3}\)
Vậy S={\(\dfrac{-3}{5};\dfrac{7}{3}\)}
c) (2-3x)(x+11)=(3x-2)(2-5x)
=(2-3x)(x+11)-(3x-2)(2-5x)=0
=(3x-2)[(x+11)-(2-5x)]=0
=(3x-2)(x+11-2+5x)=0
=(3x-2)(6x+9)=0
\(\Leftrightarrow\) 3x-2=0 hoặc 6x+9=0
\(\Leftrightarrow\) x=\(\dfrac{2}{3}\) hoặc x=\(\dfrac{-3}{2}\)
Vậy S={\(\dfrac{2}{3};\dfrac{-3}{2}\)}
d) (2x2+1)(4x-3)=(2x2+1)(x-12)
=(2x2+1)(4x-3)-(2x2+1)(x-12)=0
=(2x2+1)[(4x-3)-(x-12)=0
=(2x2+1)(4x-3-x+12)=0
=(2x2+1)(3x+9)=0
\(\Leftrightarrow\)2x2+1=0 hoặc 3x+9=0
\(\Leftrightarrow\)x=\(\dfrac{1}{2}\)hoặc x=\(\dfrac{-1}{2}\) hoặc x=-3
Vậy S={\(\dfrac{1}{2};\dfrac{-1}{2};-3\)}
e) (2x-1)2+(2-x)(2x-1)=0
=(2x-1)[(2x-1)+(2-x)=0
=(2x-1)(2x-1+2-x)=0
=(2x-1)(x+1)=0
\(\Leftrightarrow\) 2x-1=0 hoặc x+1=0
\(\Leftrightarrow\) x=\(\dfrac{-1}{2}\) hoặc x=-1
Vậy S={\(\dfrac{-1}{2}\);-1}
f)(x+2)(3-4x)=x2+4x+4
=(x+2)(3-4x)=(x+2)2
=(x+2)(3-4x)-(x+2)2=0
=(x+2)[(3-4x)-(x+2)]=0
=(x+2)(3-4x-x-2)=0
=(x+2)(-5x+1)=0
\(\Leftrightarrow\) x+2=0 hoặc -5x+1=0
\(\Leftrightarrow\) x=-2 hoặc x=\(\dfrac{1}{5}\)
Vậy S={-2;\(\dfrac{1}{5}\)}
Bạn xem lại đề nhé.
a) \(A=x^2+5y^2+2xy-4x-8y+2015\)
\(A=x^2-4x+4-2y\left(x-2\right)+y^2+2011+4y^2\)
\(A=\left(x-2\right)^2-2y\left(x-2\right)+y^2+2011+4y^2\)
\(A=\left(x-2-y\right)^2+4y^2+2011\)
Vì \(\left(x-y-2\right)^2\ge0;4y^2\ge0\)
\(\Rightarrow A_{min}=2011\)
Dấu bằng xảy ra : \(\Leftrightarrow\left\{{}\begin{matrix}x-y-2=0\\4y^2=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
a) A = 25x^2 + 10x + 1 + 3y^2 + 10
= ( 5x + 1 )^2 + 3y^21 + 10
Vậy GTNN của A là 10 khi y = 0 và 5x + 1 = 0
=> y = 0 và x = -1/5
b) Đặt x - 3 = t => x - 11 = t - 8 Ta có :
t^2 + ( t - 8 )^2 = t^2 + t^2 - 16t + 64 = 2t^2 - 16t + 64
= 2 ( t^2 - 8t + 32 ) = 2 ( t^2 - 8t + 16 + 16 ) = 2 ( t - 4 )^2 + 32
Vậy GTNN là 32 khi t = 4
=> x - 3 = 4 => x = 7
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