(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 0)x( 9 - 8 + 7 - 6 + 5 - 4 - 3 + 2 - 1)x( 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 ×9)x()= 0 (vì 1-1+2-2+3+4-7 =0 )

0

=0

Bài 1:

\(f\left(x\right)=5x-3.\)

+ \(f\left(x\right)=0\)

\(\Rightarrow5x-3=0\)

\(\Rightarrow5x=0+3\)

\(\Rightarrow5x=3\)

\(\Rightarrow x=3:5\)

\(\Rightarrow x=\frac{3}{5}\)

Vậy \(x=\frac{3}{5}.\)

+ \(f\left(x\right)=1\)

\(\Rightarrow5x-3=1\)

\(\Rightarrow5x=1+3\)

\(\Rightarrow5x=4\)

\(\Rightarrow x=4:5\)

\(\Rightarrow x=\frac{4}{5}\)

Vậy \(x=\frac{4}{5}.\)

+ \(f\left(x\right)=-2010\)

\(\Rightarrow5x-3=-2010\)

\(\Rightarrow5x=\left(-2010\right)+3\)

\(\Rightarrow5x=-2007\)

\(\Rightarrow x=\left(-2007\right):5\)

\(\Rightarrow x=-\frac{2007}{5}\)

Vậy \(x=-\frac{2007}{5}.\)

Làm tương tự với \(f\left(x\right)=2011.\)

Chúc bạn học tốt!

Bài 2: 

x=100 nên x+1=101

\(f\left(x\right)=x^8-x^7\left(x+1\right)+x^6\left(x+1\right)-x^5\left(x+1\right)\)

\(=x^8-x^8-x^7+x^7+x^6-x^6-x^5\)

\(=-x^5=-100^5\)

 X = 6 \(\times\) 8 + 6 \(\times\) 2 - 5 \(\times\) 10

X = 6 \(\times\) ( 8 + 2) - 5 \(\times\) 10

X = 6 \(\times\) 10 - 5 \(\times\) 10

X = 10 \(\times\) (6-5)

X = 10 \(\times\) 1

X = 10 

A = (90:9 + 64) \(\times\) (6 \(\times\) 6 - 4 \(\times\)9)

A = (90 : 9 + 64) \(\times\)(36 - 36)

A = (90:9 + 64) \(\times\) 0

A = 0

X = 6 x 8 + 6 x 2 - 5 x 10

X = 48 + 12 - 50

X = 10

A = ( 90 : 9 + 64 ) x ( 6 x 6 - 4 x 9 )

A = ( 10 + 64 ) x ( 36 - 36)

A = 74 x 0

A = 0

Em cam on co ah! 

hoi ma ko co ai tra loi hi.kaka

Bài 9:

Ta có: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{z}{-17}=\dfrac{-t}{-9}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{-z}{17}=\dfrac{t}{9}=-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-2\\\dfrac{-y}{3}=-2\\\dfrac{-z}{17}=-2\\\dfrac{t}{9}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\-y=-6\\-z=-34\\t=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=6\\z=34\\t=-18\end{matrix}\right.\)

Vậy: (x,y,z,t)=(-10;6;34;-18)

Bài 11:

Ta có: \(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)

\(\Leftrightarrow\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}=\dfrac{-7}{6}\)

Ta có: \(\dfrac{x}{18}=\dfrac{-7}{6}\)

\(\Leftrightarrow x=\dfrac{18\cdot\left(-7\right)}{6}=-21\)

Ta có: \(\dfrac{-98}{y}=\dfrac{-7}{6}\)

\(\Leftrightarrow y=\dfrac{-98\cdot6}{-7}=84\)

Ta có: \(\dfrac{-14}{z}=\dfrac{-7}{6}\)

\(\Leftrightarrow z=\dfrac{-14\cdot6}{-7}=12\)

Ta có: \(\dfrac{u}{-78}=\dfrac{-7}{6}\)

\(\Leftrightarrow u=\dfrac{-78\cdot\left(-7\right)}{6}=\dfrac{78\cdot7}{6}=91\)

Ta có: \(\dfrac{t}{102}=\dfrac{-7}{6}\)

\(\Leftrightarrow t=\dfrac{-7\cdot102}{6}=-7\cdot17=-119\)

Vậy: (x,y,z,t,u)=(-21;84;12;-119;91)

Nguyễn Lê Phước Thịnh giải giùm mk bài 10 đc ko ạ

\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)

\(\Leftrightarrow A=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{1}{x-2}\)

\(\Leftrightarrow A=\dfrac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}\)

\(\Rightarrow A=\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\dfrac{x-4}{x-2}\)

Ta có : \(x^2-9=0\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

Khi x = 3, thì :

\(\dfrac{3-4}{3-2}=-1\)

Khi x = -3, thì :

\(\dfrac{-3-4}{-3-2}=\dfrac{-7}{-5}=\dfrac{7}{5}\)

Vậy .............