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Toan Nguyen
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Nguyễn Hạ
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Toan Nguyen
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olm (admin@gmail.com)
29 tháng 9 2019 lúc 11:42

\(B=2x^2-4x-8=2\left(x^2-2x-4\right)\)

\(=2\left(x^2-2x+1-5\right)\)

\(=2\left[\left(x-1\right)^2-5\right]\)

\(=2\left(x-1\right)^2-10\ge-10\)

Vậy \(B_{min}=-10\Leftrightarrow x-1=0\Leftrightarrow x=1\)

\(F=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\)

Đặt \(x^2+5x+4=t\)

\(\RightarrowĐT=t\left(t+2\right)=t^2+2t+1-1\)

\(=\left(t+1\right)^2-1\ge-1\)

hay \(\left(x^2+5x+5\right)^2-1\ge-1\)

Vậy \(F_{min}=-1\Leftrightarrow x^2+5x+5=0\)

\(\Leftrightarrow x^2+5x+\frac{25}{4}-\frac{5}{4}=0\)

\(\Leftrightarrow\left(x+\frac{5}{2}\right)^2=\frac{5}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{5}{2}=\sqrt{\frac{5}{4}}\\x+\frac{5}{2}=-\sqrt{\frac{5}{4}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\frac{5}{4}}-\frac{5}{2}\\x=-\sqrt{\frac{5}{4}}-\frac{5}{2}\end{cases}}\)

olm (admin@gmail.com)
29 tháng 9 2019 lúc 11:48

\(G=4x-x^2=-\left(x^2-4x+4-4\right)\)

\(=-\left[\left(x-2\right)^2-4\right]=-\left(x-2\right)^2+4\le4\)

Vậy \(G_{max}=4\Leftrightarrow x-2=0\Leftrightarrow x=2\)

\(H=25-x-5x^2=-5\left(x^2+\frac{x}{5}-5\right)\)

\(=-5\left(x^2+2x.\frac{1}{10}+\frac{1}{100}-\frac{501}{100}\right)\)

\(=-5\left[\left(x+\frac{1}{10}\right)^2-\frac{501}{100}\right]\)

\(=-5\left(x+\frac{1}{10}\right)^2+\frac{101}{20}\le\frac{101}{2}\)

Vậy \(H_{max}=\frac{101}{2}\Leftrightarrow x+\frac{1}{10}=0\Leftrightarrow x=-\frac{1}{10}\)

Hoàng Bích Ngọc
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Nguyễn Lê Phước Thịnh
27 tháng 6 2023 lúc 22:21

2:

a: A(x)=0

=>5x-10-2x-6=0

=>3x-16=0

=>x=16/3

b: B(x)=0

=>5x^2-125=0

=>x^2-25=0

=>x=5 hoặc x=-5

c: C(x)=0

=>2x^2-x-3=0

=>2x^2-3x+2x-3=0

=>(2x-3)(x+1)=0

=>x=3/2 hoặc x=-1

Yến Nhi Sky M-tp
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Trịnh Ngọc Hân
24 tháng 7 2018 lúc 19:12

a) \(3x^2-3y^2-x-y\)

\(\Leftrightarrow3\left(x^2-y^2\right)-x-y\)

\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)

\(\Leftrightarrow3\left(x-y\right)\)

d) \(3x^2-7x+4\)

\(\Leftrightarrow3x^2-7x+7-3\)

\(\Leftrightarrow\left(3x^2-3\right)-\left(7x-7\right)\)

\(\Leftrightarrow3\left(x^2-1\right)-7\left(x-1\right)\)

\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(3\left(x+1\right)-7\right)\)

\(\Leftrightarrow\left(x+1\right)\left(3x-6\right)\)

e) \(-2x^2+3x-1\)

\(\Leftrightarrow\left(-2x^2-1^2\right)+3x\)

\(\Leftrightarrow\left(-2x-1\right)\left(-2x+1\right)+3x\)

f) \(x^2+2xy+y^2-2x-2y\)

\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)

\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)

k) \(2x^2+5x+3\)

\(\Leftrightarrow2x^2+2x+3x+3\)

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)\)

\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)\)

l) \(x^2-2x-y^2+1\)

\(\Leftrightarrow\left(x^2-2x+1\right)-y^2\)

\(\Leftrightarrow\left(x-1\right)^2-y^2\)

\(\Leftrightarrow\left(x-1-y\right)\left(x-1+y\right)\)

Trịnh Ngọc Hân
24 tháng 7 2018 lúc 19:12

a) \(3x^2-3y^2-x-y\)

\(\Leftrightarrow3\left(x^2-y^2\right)-x-y\)

\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)

\(\Leftrightarrow3\left(x-y\right)\)

d) \(3x^2-7x+4\)

\(\Leftrightarrow3x^2-7x+7-3\)

\(\Leftrightarrow\left(3x^2-3\right)-\left(7x-7\right)\)

\(\Leftrightarrow3\left(x^2-1\right)-7\left(x-1\right)\)

\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(3\left(x+1\right)-7\right)\)

\(\Leftrightarrow\left(x+1\right)\left(3x-6\right)\)

e) \(-2x^2+3x-1\)

\(\Leftrightarrow\left(-2x^2-1^2\right)+3x\)

\(\Leftrightarrow\left(-2x-1\right)\left(-2x+1\right)+3x\)

f) \(x^2+2xy+y^2-2x-2y\)

\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)

\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)

k) \(2x^2+5x+3\)

\(\Leftrightarrow2x^2+2x+3x+3\)

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)\)

\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)\)

l) \(x^2-2x-y^2+1\)

\(\Leftrightarrow\left(x^2-2x+1\right)-y^2\)

\(\Leftrightarrow\left(x-1\right)^2-y^2\)

\(\Leftrightarrow\left(x-1-y\right)\left(x-1+y\right)\)

♥๖Lan_Phương_cute#✖#girl...
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Ngô Chi Lan
29 tháng 8 2020 lúc 13:29

It's khai triển :)

a) \(\left(5x-x^2\right)\left(5x+x^2\right)=25x^2-x^4\)

b) \(\left(2x-y\right)\left(4x^2+2xy+y^2\right)=8x^3-y^3\)

c) \(\left(x+3\right)\left(x^2-3x+9\right)=x^3-27\)

d) \(-x^3+3x^2-3x+1=\left(1-x\right)^3\)

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Ngô Chi Lan
29 tháng 8 2020 lúc 13:31

e) \(x^2-2x+9=\left(x-1\right)^2+8??\) ko ra gì cả-.-

g) \(\left(x+1\right)\left(x-1\right)=x^2-1\)

h) \(\left(x-2y\right)\left(x+2y\right)=x^2-4y^2\)

i) \(25a^2+4b^2-20ab=\left(5a-2b\right)^2\)

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help me
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help me
12 tháng 7 2016 lúc 20:59

nhanh lên các bạn nhé mai mình đi học rồi

Nguyen Thi Thu Ha
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Nguyễn Kim Chi
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Võ Đông Anh Tuấn
20 tháng 6 2017 lúc 9:43

a ) \(x^2-x+1\)

\(\Leftrightarrow\left(x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right)+\dfrac{3}{4}\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Ta có : \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Vậy GTNN là \(\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}.\)