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a) Ta có: \(\left(2x+1\right)^2+\left(1-x\right)3x\le\left(x+2\right)^2\)
\(\Leftrightarrow x^2+4x+4\ge4x^2+4x+1+3x-3x^2\)
\(\Leftrightarrow x^2+4x+4\ge x^2+7x+1\)
\(\Leftrightarrow3\ge3x\)
\(\Rightarrow x\le1\)
b) Ta có: \(\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)
\(\Leftrightarrow x^2-16\ge x^2+6x+9+5\)
\(\Leftrightarrow6x\le-30\)
\(\Leftrightarrow x\le-5\)
a) ( 2x + 1 )2 + ( 1 - x )3x ≤ ( x + 2 )2
<=> 4x2 + 4x + 1 + 3x - 3x2 ≤ x2 + 4x + 4
<=> 4x2 + 4x + 3x - 3x2 - x2 - 4x ≤ 4 - 1
<=> 3x ≤ 3
<=> x ≤ 1
b) ( x - 4 )( x + 4 ) ≥ ( x + 3 )2 + 5
<=> x2 - 16 ≥ x2 + 6x + 9 + 5
<=> x2 - x2 - 6x ≥ 9 + 5 + 16
<=> -6x ≥ 30
<=> x ≤ -5
x3 - ( a + b + c )x2 + ( ab + bc + ca )x = abc
<=> x3 - ax2 - bx2 - cx2 + abx + bcx + cax - abc = 0
<=> x3 - ax2 - bx2 + abx - cx2 + bcx + cax - abc = 0
<=> x ( x2 - ax - bx + ab ) - c ( x2 - bx - ax + ab ) = 0
<=> ( x - c ) ( x2 - ax - bx + ab ) = 0
<=> ( x - c ) [ x ( x - b ) - a ( x - b ) ] = 0
<=> ( x - c ) ( x - a ) ( x - b ) = 0
<=>\(\hept{\begin{cases}x-c=0\\x-a=0\\x-b=0\end{cases}}\) <=> a = b = c = x
2/ Áp dụng BĐT Bunhiacopxki \(\left(ax+by\right)^2\le\left(a^2+b^2\right)\left(x^2+y^2\right)\)
\(\Leftrightarrow a^2x^2+b^2y^2+2abxy\le a^2x^2+a^2y^2+b^2x^2+b^2y^2\)
\(\Leftrightarrow bx^2+ay^2-2abxy\ge0\)
\(\Leftrightarrow\left(bx-ay\right)^2\ge0\)(đúng) Dấu "=" xảy ra khi x/a=y/b
Ta có: \(\left(x+4y\right)^2\le\left(1^2+2^2\right)\left(x^2+4y^2\right)=5\left(x^2+4y^2\right)\)
Mà a + 4b = 1
\(\Rightarrow x^2+4y^2\ge\frac{1}{5}\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\frac{1}{x}=\frac{2}{2y}=\frac{1}{y}\\x+4y=1\end{cases}}\Rightarrow x=y=\frac{1}{5}\)
Bài 1:
a) Ta có: \(VT=\frac{-u^2+3u-2}{\left(u+2\right)\left(u-1\right)}\)
\(=\frac{-\left(u^2-3u+2\right)}{\left(u+2\right)\left(u-1\right)}\)
\(=\frac{-\left(n^2-u-2u+2\right)}{\left(u+2\right)\left(u-1\right)}\)
\(=\frac{-\left[u\left(u-1\right)-2\left(u-1\right)\right]}{\left(u+2\right)\left(u-1\right)}\)
\(=\frac{-\left(u-1\right)\left(u-2\right)}{\left(u+2\right)\left(u-1\right)}\)
\(=\frac{2-u}{u+2}\)(1)
Ta có: \(VP=\frac{u^2-4u+4}{4-u^2}\)
\(=\frac{\left(u-2\right)^2}{-\left(u-2\right)\left(u+2\right)}\)
\(=\frac{-\left(u-2\right)}{u+2}\)
\(=\frac{2-u}{u+2}\)(2)
Từ (1) và (2) suy ra \(\frac{-u^2+3u-2}{\left(u+2\right)\left(u-1\right)}=\frac{u^2-4u+4}{4-u^2}\)
b) Ta có: \(VT=\frac{v^3+27}{v^2-3v+9}\)
\(=\frac{\left(v+3\right)\left(v^3-3u+9\right)}{v^2-3u+9}\)
\(=v+3=VP\)(đpcm)
Bài 2:
a) Ta có: \(\frac{3x^2-2x-5}{M}=\frac{3x-5}{2x-3}\)
\(\Leftrightarrow\frac{3x^2-5x+3x-5}{M}=\frac{3x-5}{2x-3}\)
\(\Leftrightarrow\frac{x\left(3x-5\right)+\left(3x-5\right)}{M}=\frac{3x-5}{2x-3}\)
\(\Leftrightarrow\frac{\left(3x-5\right)\left(x+1\right)}{M}=\frac{3x-5}{2x-3}\)
\(\Leftrightarrow M=\frac{\left(3x-5\right)\left(x+1\right)\left(2x-3\right)}{3x-5}\)
\(\Leftrightarrow M=\left(x+1\right)\left(2x-3\right)\)
\(\Leftrightarrow M=2x^2-3x+2x-3\)
hay \(M=2x^2-x-3\)
Vậy: \(M=2x^2-x-3\)
b) Ta có: \(\frac{2x^2+3x-2}{x^2-4}=\frac{M}{x^2-4x+4}\)
\(\Leftrightarrow\frac{2x^2+4x-x-2}{\left(x-2\right)\left(x+2\right)}=\frac{M}{\left(x-2\right)^2}\)
\(\Leftrightarrow\frac{2x\left(x+2\right)-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{M}{\left(x-2\right)^2}\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(2x-1\right)}{\left(x+2\right)\left(x-2\right)}=\frac{M}{\left(x-2\right)^2}\)
\(\Leftrightarrow\frac{M}{\left(x-2\right)^2}=\frac{2x-1}{x-2}\)
\(\Leftrightarrow M=\frac{\left(2x-1\right)\left(x-2\right)^2}{\left(x-2\right)}\)
\(\Leftrightarrow M=\left(2x-1\right)\left(x-2\right)\)
\(\Leftrightarrow M=2x^2-4x-x+2\)
hay \(M=2x^2-5x+2\)
Vậy: \(M=2x^2-5x+2\)
Bài 3:
a) Ta có: \(\frac{x+1}{N}=\frac{x^2-2x+4}{x^3+8}\)
\(\Leftrightarrow\frac{x+1}{N}=\frac{x^2-2x+4}{\left(x+2\right)\left(x^2-2x+4\right)}\)
\(\Leftrightarrow\frac{x+1}{N}=\frac{1}{x+2}\)
\(\Leftrightarrow N=\left(x+1\right)\left(x+2\right)\)
hay \(N=x^2+3x+2\)
Vậy: \(N=x^2+3x+2\)
n) Ta có: \(\frac{\left(x-3\right)\cdot N}{3+x}=\frac{2x^3-8x^2-6x+36}{2+x}\)
\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=\frac{2x^3+4x^2-12x^2-24x+18x+36}{x+2}\)
\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{\left(x+3\right)}=\frac{2x^2\left(x+2\right)-12x\left(x+2\right)+18\left(x+2\right)}{x+2}\)
\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=\frac{\left(x+2\right)\left(2x^2-12x+18\right)}{x+2}\)
\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=2x^2-12x+18\)
\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=2x^2-6x-6x+18=2x\left(x-3\right)-6\left(x-3\right)=2\cdot\left(x-3\right)^2\)
\(\Leftrightarrow N\cdot\left(x-3\right)=\frac{2\left(x-3\right)^2}{x+3}\)
\(\Leftrightarrow N=\frac{2\left(x-3\right)^2}{x+3}:\left(x-3\right)=\frac{2\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}\)
\(\Leftrightarrow N=\frac{2\left(x-3\right)}{x+3}\)
hay \(N=\frac{2x-6}{x+3}\)
Vậy: \(N=\frac{2x-6}{x+3}\)
a) x3+ 6x2+12x+8
=(x+2)3
b)x3-3x2+3x-1
=(x-1)3
c)1-9x+27x2-27x3
=(1-3x)3
d)x3 +\(\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}\)
=(x+\(\frac{1}{2}\))3 ( phần này mik là khác đầu bài bạn đi 1 chút nhưng mik tôn trọng ý kiến của bạn hơn nên mik nghĩ mik làm sai)
e) 27x3-54x2y+36xy2-8y3
=(3x-2y)2
a) x3 + 6x2 + 12x + 8
= (x^3+2^3)+6x.(x+2)
= (x+2).(x^2-2x+4)+6x(x+2)
= (x+2).(x^2+4x+4)
b) x3 - 3x2 + 3x - 1
= (x^3-1) -3x.(x-1)
= (x-1).(x^2+x+1) - 3x(x-1)
= (x-1).(x^2-2x+1)
Câu d ko hiểu đề :v
e) 27x3- 54 x2y + 36 xy2 - 8y3
= (27x^3-8y^3)-(54x^2y+36xy^2)
= (3x-2y).(9x^2+6xy+4y^2)-18xy(3x-2y)
= (3x-2y).(9x^2-12xy+4y^2)
Thế nhé :)
Khó thế mình không biết làm!
Minh cung k biet lam vi mik hoc lop 5