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= \(4x^2\)+\(20x\)+\(25\)+\(6x^2\)- \(8x\)- \(x^2\)-\(22\)
=\(9x^2\)+\(12x\)+\(3\)
=\(9x^2\)+\(12x\)+\(3\)
=\(9x^2\)+\(12x\)+\(4\)-\(1\)
=(\(3x\)+\(2\))2-\(1\)
vì (\(3x\)+\(2\))2 >-0
=>.................-\(1\)>-(-1)
(>- là > hoặc =)
=> GTNN của M= -1 khi và chỉ khi \(3x\)+\(2\)=\(0\)
..................................
Đặt x2-2x+1=t, ta có:
\(A=\left(t-1\right)\left(t+1\right)=t^2-1=\left(x^2-2x+1\right)^2-1\ge-1\)
Dấu "=" xảy ra khi \(x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)
Đặt \(\left(x^2-2x\right)\left(x^2-2x=2\right)=k.\left(k+2\right)=A\)
\(\Rightarrow A=k.\left(k+2\right)=k^2+2k\)
\(\Rightarrow A=k^2+k+k+1-1=k\left(k+1\right)+\left(k+1\right)-1\)
\(\Rightarrow A=\left(k+1\right)^2-1\)
\(\Rightarrow A=\left(x^2-2x+1\right)^2-1\)
\(\Rightarrow A=\left(x^2-x-x+1\right)^2-1=\left[x.\left(x-1\right)-\left(x-1\right)\right]^2-1\)
\(\Rightarrow A=\left(x-1\right)^2-1\ge-1\)
( Dấu "=" xảy ra <=> x=1 )
a) đK: \(x\ne0;2\)
B = \(\dfrac{3x-4}{x\left(x-2\right)}.\dfrac{x\left(x-2\right)}{x^2-4-x^2}=\dfrac{3x-4}{-4}=\dfrac{4-3x}{4}\) \(\dfrac{x-4+2x}{x\left(x-2\right)}:\dfrac{\left(x-2\right)\left(x+2\right)-x^2}{x\left(x-2\right)}\)
= \(\dfrac{3x-4}{x\left(x-2\right)}.\dfrac{x\left(x-2\right)}{x^2-4-x^2}=\dfrac{4-3x}{4}\)
b) Thay x = -2 (TMDK) vào B, ta có:
\(B=\dfrac{4-3.\left(-2\right)}{4}=\dfrac{4+6}{4}=\dfrac{5}{2}\)
c) Để \(\left|B\right|-2x=5\)
<=> \(\left|\dfrac{4-3x}{4}\right|-2x=5\)
TH1: \(x\le\dfrac{4}{3}\)
<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{4-3x}{4}\)
PT <=> \(\dfrac{4-3x}{4}-2x=5\)
<=> \(\dfrac{4-3x-8x}{4}=5\)
<=> \(4-11x=20\)
<=> x = \(\dfrac{-16}{11}\) (Tm)
TH2: \(x>\dfrac{4}{3}\)
<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{3x-4}{4}\)
PT <=> \(\dfrac{3x-4}{4}-2x=5\)
<=> \(\dfrac{3x-4-8x}{4}=5\)
<=> \(-5x-4=20\)
<=> \(x=\dfrac{-24}{5}\left(l\right)\)
d) Xét (2-x)B = \(\dfrac{\left(2-x\right)\left(4-3x\right)}{4}\) = \(\dfrac{3x^2-10x+8}{4}\)
= \(\dfrac{3\left(x-\dfrac{5}{3}\right)^2-\dfrac{1}{3}}{4}\)
Mà \(3\left(x-\dfrac{5}{3}\right)^2\ge\) 0
=> (2-x)B \(\ge\dfrac{\dfrac{-1}{3}}{4}=\dfrac{-1}{12}\)
Dấu "=" <=> x = \(\dfrac{5}{3}\left(tm\right)\)
e) Số nguyên âm lớn nhất là -1
Để B = -1
<=> \(\dfrac{4-3x}{4}=-1\)
<=> 4 - 3x = -4
<=> \(x=\dfrac{8}{3}\left(tm\right)\)
g)
TH1: \(x\le\dfrac{4}{3}\)
<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{4-3x}{4}\)
BDT <=> \(\dfrac{4-3x}{4}< 2x-4\)
<=> \(4-3x< 8x-16\)
<=> \(x>\dfrac{20}{11}\left(l\right)\)
TH2: \(x>\dfrac{4}{3}\)
<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{3x-4}{4}\)
BDT <=> \(\dfrac{3x-4}{4}< 2x-4\)
<=> \(3x-4< 8x-16\)
<=> x > \(\dfrac{12}{5}\)
KHDK: \(x>\dfrac{12}{5}\)
\(A=\left(x-1\right)\left(2x-1\right)\left(2x^2-3x-1\right)+2017\)
\(=\left(2x^2-3x+1\right)\left(2x^2-3x-1\right)+2017\)
\(=\left(2x^2-3x\right)^2-1+2017\)
\(=\left(2x^2-3x\right)^2+2016\ge2016\)
\(\Leftrightarrow2x^2-3x=0\Leftrightarrow x\left(2x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)
Vậy \(A_{min}=2016\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)
a, \(A=\left(\frac{4}{2x+1}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\left(\frac{4\left(x^2+1\right)}{\left(2x+1\right)\left(x^2+1\right)}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\left(\frac{4x^2+4+4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\frac{\left(2x+1\right)^2}{\left(x^2+1\right)\left(2x+1\right)}\frac{x^2+1}{x^2+2}=\frac{2x+1}{x^2+2}\)
Gọi 1/4 số a là 0,25 . Ta có :
a . 3 - a . 0,25 = 147,07
a . (3 - 0,25) = 147,07 ( 1 số nhân 1 hiệu )
a . 2,75 = 147,07
a = 147,07 : 2,75
a = 53,48
mình nha