a: f(0)=5

f(-3)=3

b: y=6

=>2/3*x=1

hay x=3/2

3/2 nha

Đề bài yêu cầu gì?

\(\left(x+2\right)\left(x-3\right)\left(x-6\right)< 0\)

Suy ra phải có ít nhất 1 số âm

Lại có: \(x-6< x-3< x+2\)

nên \(\hept{\begin{cases}x-6< 0\\x-3>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 6\\x>3\end{cases}}\Leftrightarrow3< x< 6\)

giải chi tiết nha mn

CM :nghiệm à bn

x⁶+x⁵+x⁴+x³+x²+x+1=0

(x⁷-1)/(x-1)=0

với x=1 thì 1+1+1+1+1+1+1=0 vô lí 

với x khác 1 thì:

x⁷-1/x-1=0

x⁷-1=0

x⁷=-1

x=-1

tìm x nha bạn

\(\left(-3x+2\right)-\left(5-3x\right)=-3\)

\(\Rightarrow-3x+2-5+3x=-3\)

\(\Rightarrow-3x+3x=-3+5-2\)

\(\Rightarrow0x=0\Rightarrow x\in Z\)

\(3+x-\left(3x-1\right)=6-2x\)

\(\Rightarrow3+x-3x+1=6-2x\)

\(\Rightarrow x-3x+2x=6-1-3\)

\(\Rightarrow0x=2\left(loại\right)\)

\(\left(x-5\right)\left(3x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-\frac{4}{3}\end{cases}}}\)

\(7x\left(2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}7x=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)

\(\left(3x-1\right)2x=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=0\end{cases}}}\)

khó hiểu quá

bạn ghi bằng số luôn đừng ghi phần

\(\left(-3x+2\right)-\left(5-3x\right)=-3\)

\(\Rightarrow-3x+2-5+3x=-3\)

\(\Rightarrow-3=-3\)

\(\forall x\in Z\)

\(3+x-\left(3x-1\right)=6-2x\)

\(\Rightarrow2=6\left(vl\right)\)

\(\left(x-5\right)\left(3x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\3x+4=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=\frac{4}{3}\end{cases}}\)

\(7x\left(2x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7x=0\\2x-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}\)

\(\left(3x-1\right)2x=0\)

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\2x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=0\end{cases}}\)

a) x=3/2

b)x=-1

c) x=5

d) x= 5/2

/5x-4/=/x+2/

\(\orbr{\begin{cases}5x-4=x+2\\5x-4=-x+2\end{cases}}suyra\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{2}\end{cases}}\)

vậy x=3/2 hoặc x=1/2

Ta có : \(\left|5x-4\right|=\left|x+2\right|\)

\(\Leftrightarrow\orbr{\begin{cases}5x-4=x+2\\5x-4=-x-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x-x=2+4\\5x+x=-2+4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x=6\\6x=2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{3}\end{cases}}\)

b) \(\left|2x-3\right|-\left|3x+2\right|=0\)

\(\Rightarrow\orbr{\begin{cases}2x-3=3x+2\\2x-3=-3x-2\end{cases}\Rightarrow\orbr{\begin{cases}2x-3x=2+3\\2x+3x=-2+3\end{cases}\Rightarrow}\orbr{\begin{cases}-x=5\\5x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x=\frac{1}{5}\end{cases}}}\)

c)/2+3x/=/4x-3/

\(\Rightarrow\orbr{\begin{cases}2+3x=4x-3\\2+3x=-\left(4x-3\right)\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x-4x=-3-2\\3x+4x=3-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-x=-5\\7x=1\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{7}\end{cases}}}\)

d)/7x+1/-/5x+6|=0

\(\Rightarrow\left|7x+1\right|=\left|5x+6\right|\)

\(\Rightarrow\orbr{\begin{cases}7x+1=5x+6\\7x+1=-\left(5x+6\right)\end{cases}\Rightarrow\orbr{\begin{cases}7x-5x=6-1\\7x+1=-5x-6\end{cases}\Rightarrow}\orbr{\begin{cases}2x=5\\7x+5x=-6-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{7}{12}\end{cases}}}\)

Toán mik ghi nhầm ko phải ta

em ghi nhầm môn em có thể đăng lại ko em

y) \(x^2-x-6=0\\ \Leftrightarrow\left(x-3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{-2;3\right\}\) là nghiệm của pt.

z) \(3x^2-5x-8=0\\ \Leftrightarrow\left(3x-8\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-1\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{8}{3};-1\right\}\) là nghiệm của pt.

j) \(25x^2-4=0\\ \Leftrightarrow\left(5x-2\right)\left(5x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{-2}{5}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{2}{5};\dfrac{-2}{5}\right\}\) là nghiệm của pt.

r) \(2\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

Vậy \(x\in\left\{-3;2\right\}\) là nghiệm của pt.

u) \(x^3-3x^2-x+3=0\\ \Leftrightarrow x^2\left(x-3\right)-\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x^2-1\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\\x=-1\end{matrix}\right.\)

Vậy \(x\in\left\{-1;1;3\right\}\) là nghiệm của pt.

y: Ta có: \(x^2-x-6=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

z: Ta có: \(3x^2-5x-8=0\)

\(\Leftrightarrow\left(3x-8\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-1\end{matrix}\right.\)

j: Ta có: \(25x^2-4=0\)

\(\Leftrightarrow\left(5x-2\right)\left(5x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

r: Ta có: \(2\left(x+3\right)-x^2-3x=0\)

\(\Leftrightarrow\left(x+3\right)\left(2-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

u: Ta có: \(x^3-3x^2-x+3=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\\x=-1\end{matrix}\right.\)