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a) Ta thấy:
\(\left(x+4\right)\left(x-4\right)=x\left(x-\frac{2}{3}\right)\)
\(\Rightarrow\left(x^2-4x\right)+\left(4x-16\right)=x^2-\frac{2}{3}x\)
\(\Rightarrow\left(x^2-16\right)-\left(4x-4x\right)=x^2-\frac{2}{3}x\)
\(\Rightarrow x^2-16-0=x^2-\frac{2}{3}x\)
\(\Rightarrow x^2-16=x^2-\frac{2}{3}x\)
\(\Rightarrow16=\frac{2}{3}x\) ( do có cùng hiệu và cùng số bị trừ )
\(\Rightarrow x=16:\frac{2}{3}\)
\(\Rightarrow x=24\)
Vậy x = 24
b.) x^3-x^2-2x=0
x(x^2-x-2)=0
x(x^2-2x+x-2)=0
x(x(x-2)+x-2)=0
x(x-2)(x+1)=0
suy ra x=0 hoặc x-2=0 hoặc x+1=0
vậy x=0 hoặc x=2 hoặc x=-1
hình như câu c đề phải là (x+4)/120 thì phải đó bạn
c.)(x+4)/120+(x+8)/116=(x+5)/119+(x+7)/117
(x+4)/120+(x+8)/116-(x+5)/119-(x+7)/117=0
(x+4)/120+1+(x+8)/116+1-(x+5)/119-1-(x+7)/117-1=0
(x+4)/120+1+(x+8)/116+1-((x+5)/119+1)-((x+7)/117+1)=0
(x+124)/120+(x+124)/116-(x+124)/119-(x+124)/117=0
(x+124)(1/120+1/116-1/119-1/117)=0
suy ra x+124=0
x=-124
Câu 1: \(x^2+\frac{1}{x^2}-4x-\frac{4}{x}+6=0\)
\(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)-4\left(x+\frac{1}{x}\right)+6=0\)
\(\text{Đặt a = }x+\frac{1}{x}\)
\(\Rightarrow a^2=\left(x+\frac{1}{x}\right)^2=x^2+2.x.\frac{1}{x}+\left(\frac{1}{x}\right)^2=x^2+2+\frac{1}{x^2}\)
\(\Rightarrow x^2+\frac{1}{x^2}=a^2-2\)
Thay vào phương trình ta có:
\(\left(a^2-2\right)-4a+6=0\)
\(\Leftrightarrow a^2-2-4a+4=0\)
\(\Leftrightarrow a^2-4a+4=0\)
\(\Leftrightarrow\left(a-2\right)^2=0\)
\(\Leftrightarrow a-2=0\)
\(\Rightarrow x+\frac{1}{x}-2=0\)\(ĐKXĐ:x\ne0\)
\(\Leftrightarrow\frac{x^2+1-2x}{x}=0\)
\(\Leftrightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy x=1
a) \(-7x^2+10x-2016=-7\left(x^2-\frac{10x}{7}\right)-2016=-7\left(x^2-2.x.\frac{5}{7}+\frac{25}{49}\right)+\frac{25}{49}.7-2016=-7\left(x-\frac{5}{7}\right)^2-\frac{14087}{7}\le-\frac{14087}{7}\)Vậy Max = \(-\frac{14087}{7}\Leftrightarrow x=\frac{5}{7}\)
b) \(\frac{x+5}{11}+\frac{x+2010}{6}\ge\frac{x-1}{2017}+\frac{x+6}{2010}\)
\(\Leftrightarrow\frac{x}{2011}+\frac{x}{6}+\frac{5}{2011}+335\ge\frac{x}{2017}+\frac{x}{2010}-\frac{1}{2017}+\frac{1}{335}\)
\(\Leftrightarrow x\left(\frac{1}{2011}+\frac{1}{6}-\frac{1}{2017}-\frac{1}{2010}\right)\ge\frac{1}{335}-\frac{1}{2017}-\frac{5}{2011}-335\)
\(\Leftrightarrow\frac{677389259}{4076467935}x\ge\frac{-455205582048}{1358822645}\) \(\Leftrightarrow x\ge-2016\)
Câu b) còn cách khác nữa bạn nhé. Mình làm cách này "xù" quá ^^
ĐKXĐ:\(x\ne\pm2;x\ne-3;x\ne0\)
\(P=1+\frac{x-3}{x^2+5x+6}\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left(\frac{2}{x-2}-\frac{x}{x^2-4}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\cdot\frac{2x+4-x-x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\frac{8\left(x-3\right)}{\left(x+2\right)^2\left(x+3\right)\left(x-2\right)}\)
Đề sai à ??
1. đặt b + c - a = x, a + c - b = y , a + b - c = z thì x,y,z > 0
theo bất đẳng thức ( x + y ) ( y + z ) ( x + z ) \(\ge\)8xyz ( tự chứng minh ) , ta có :
2a . 2b . 2c \(\ge\)8 ( b + c - a ) ( a + c - b ) ( a + b - c )
\(\Rightarrow\)abc \(\ge\)( b + c - a ) ( a + c - b ) ( a + b - c )
Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c
Ta có a + b > c, b + c > a, a + c > b
Xét \(\frac{1}{a+c}+\frac{1}{b+c}>\frac{1}{a+c+b}+\frac{1}{b+c+a}=\frac{2}{a+b+c}>\frac{2}{a+b+a+b}=\frac{1}{a+b}\)
tương tự : \(\frac{1}{a+b}+\frac{1}{a+c}>\frac{1}{b+c},\frac{1}{a+b}+\frac{1}{b+c}>\frac{1}{a+c}\)
vậy ...
1) Ta có pt : \(4x^2+\frac{1}{x^2}=8x+\frac{4}{x}\)
\(\Leftrightarrow4x^2+4+\frac{1}{x^2}=8x+4+\frac{4}{x}\)
\(\Leftrightarrow\left(2x+\frac{1}{x}\right)^2=4\left(2x+\frac{1}{x}\right)+4\)
\(\Leftrightarrow\left(2x+\frac{1}{x}\right)^2-4\left(2x+\frac{1}{x}\right)+4=8\)
\(\Leftrightarrow\left(2x+\frac{1}{x}-2\right)^2=8\)
Đến đây dễ rồi nhé, chia 2 TH.