x (x+1) + x (x-5) - 5( x+1)
P/s: Giups mình với các bẹn ưiii❤
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\(\left(x:2,2\right)\times\dfrac{1}{6}=\dfrac{-3}{8}\times\left(0,5-1\dfrac{3}{5}\right)\)
\(\Rightarrow\left(x:2,2\right)\times\dfrac{1}{6}=\dfrac{-3}{8}\times\left(\dfrac{1}{2}-\dfrac{8}{5}\right)\)
\(\Rightarrow\left(x:2,2\right)\times\dfrac{1}{6}=\dfrac{-3}{8}\times\dfrac{11}{10}\)
\(\Rightarrow\left(x:2,2\right)\times\dfrac{1}{6}=\dfrac{33}{80}\)
\(\Rightarrow x:2,2=\dfrac{33}{80}:\dfrac{1}{6}\)
\(\Rightarrow x:2,2=\dfrac{99}{40}\)
\(\Rightarrow x=\dfrac{99}{40}\times2,2\)
\(\Rightarrow x=\dfrac{1089}{200}\)
=>(x:2,2)*1/6=-3/8(1/2-8/5)=33/80
=>x:2,2=99/40
=>x=1089/200
\(\left(x+2\right)\left(x^2+2x-9\right)\)
\(=x^3+2x^2-9x+2x^2+4x-18\)
\(=x^3+4x^2-5x-18\)
\(\left(x^{2y}-6\right)\left(x^2-5\right)\)
\(=x^{4y}-5x^{2y}-6x^2+30\)
\(\left(x+y\right)\left(xy-4+y\right)\)
\(=x^2y-4x+xy+xy^2-4y+y^2\)
câu còn lại tương tự nha
`Answer:`
\(\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{9}\right)+\left(x+\frac{1}{27}\right)+...+\left(x+\frac{1}{729}\right)=\frac{4209}{729}\)
\(\Leftrightarrow\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{3^2}\right)+\left(x+\frac{1}{3^3}\right)+...+\left(x+\frac{1}{3^6}\right)=\frac{4209}{729}\)
\(\Leftrightarrow6x+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^6}\right)=\frac{4209}{729}\text{(*)}\)
Đặt \(N=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^6}\)
\(\Leftrightarrow3N=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
\(\Leftrightarrow3N-N=\left(1+\frac{1}{3}+\frac{1}{3^2}+..+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^6}\right)\)
\(\Leftrightarrow2N=1-\frac{1}{3^6}\)
\(\Leftrightarrow2N=\frac{728}{729}\)
\(\Leftrightarrow N=\frac{364}{729}\)
\(\text{(*)}\Leftrightarrow6x+\frac{364}{729}=\frac{4209}{729}\)
\(\Leftrightarrow6x=\frac{3845}{729}\)
\(\Leftrightarrow x=\frac{3845}{4374}\)
S = 1 x 3 + 3 x 5 + 5 x 7 + 7 x 9 + 9 x 11 + 11 13 + 13 x 15 + 15 x 17 + 17 x 19 + 19 x 21 + 21 x 23 + 23 x 25
Cho đa thức f(x)=x2-7x+6 và đa thức g(x)2x2+3x-5 .so sánh f(1)với g(1)
Giups mình với . mìnhđang gấp
chúc mừng bạn đã mất căn bản
f(1) = 0
g(1) = 0
Vậy f(1) = g(1)
EZ
\(x\left(x+1\right)+x\left(x-5\right)-5\left(x+1\right)\)
= \(\left(x+1\right)\left(x-5\right)+x\left(x-5\right)\)
= \(\left(x-5\right)\left(x+1+x\right)=\left(x-5\right)\left(2x+1\right)\)