b) Giải:
Ta có: \(4x+3⋮x-2\)

\(\Rightarrow4x-8+11⋮x-2\)

\(\Rightarrow4\left(x-2\right)+11⋮x-2\)

\(\Rightarrow11⋮x-2\)

\(\Rightarrow x-2\in\left\{1;-1;11;-11\right\}\)

\(\left[\begin{matrix}x-2=1\\x-2=-1\\x-2=11\\x-2=-11\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=3\\x=1\\x=13\\x=-9\end{matrix}\right.\)

Vậy \(x\in\left\{3;1;13;-9\right\}\)

b.Ta có:(4x+3)=4x-4.2+8+3

=4(x-2)+11

Để(4x+3)chia hết cho (x-2)

#11chia hết cho (x-2)(#là khi và chỉ khi nhế!)

#x-2€ Ư(11)={±1;±11}

#x€{3;1;13;-9}

Vậy x€{3;1;13;-9}

a) \(\left|x\right|< 1\Rightarrow-1< x< 1\Rightarrow x=0\)

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

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

c) \(\left|x+2\right|=\left|12-10\right|\)

\(\Leftrightarrow\left|x+2\right|=2\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=0\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}2x-2\ge0\\\left[{}\begin{matrix}x+3=2x-2\\x+3=\left(-2x\right)+2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x\ge2\\\left[{}\begin{matrix}x-2x=-2-3\\x-\left(-2x\right)=2-3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\left[{}\begin{matrix}-x=-5\\3x=-1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\\left[{}\begin{matrix}x=5\left(tm\right)\\x=\dfrac{-1}{3}\end{matrix}\right.\end{matrix}\right.\)

Vì \(\dfrac{-1}{3}< 1\) nên \(x=5\) thỏa mãn đề bài.

e) \(\left|x+1\right|>4\)

\(\Rightarrow\left[{}\begin{matrix}x+1>4\\x+1< 4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x>3\\x< 3\end{matrix}\right.\)

f) \(\left|x-3\right|=\left|2x-1\right|\)

(cho thời gian suy nghĩ, mình chưa làm dạng này bao giờ)

g) \(\left|2x-1\right|-1+2x=0\)

\(\Rightarrow\left|2x-1\right|=-2x+1\)

Mà \(\left|2x-1\right|=\left|-2x+1\right|\)

\(\Rightarrow\left|-2x+1\right|=-2x+1\)

\(\Rightarrow-2x+1\ge0\)

\(\Rightarrow-2x\ge-1\)

\(\Rightarrow x\ge\dfrac{-1}{-2}=\dfrac{1}{2}\)

h) \(\left|3-2x\right|=2x-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3\ge0\\\left[{}\begin{matrix}3-2x=2x-3\\3-2x=-2x+3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x\ge3\\\left[{}\begin{matrix}3+3=2x+2x\\3-3=-2x+2x\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\left[{}\begin{matrix}6=4x\\0=0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\left[{}\begin{matrix}x=\dfrac{3}{2}\\0=0\end{matrix}\right.\end{matrix}\right.\)

Vì \(0=0\) luôn đúng nên ta có \(x=\dfrac{3}{2}\)

j) \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|=5x\)

(đầu hàng)

\(a,\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\) \(=x^3+1-x^3+1=2\)

\(b,x\left(x-4\right)\left(x+4\right)-\left(x^2+1\right)\left(x^2-1\right)\)

\(=x\left(x^2-16\right)-\left(x^4-1\right)=x^3-16x-x^4+1\) \(c,\left(x-3\right)\left(x+3\right)-\left(x+1\right)^2\)

\(=x^2-9-x^2-2x-1=-2x-10\)

\(d,\left(4x-3\right)\left(4x+3\right)-16x^2\)

\(=16x^2-9-16x^2=-9\)

\(e,\left(x+4\right)\left(x^2-4x+16\right)-x^3=x^3+64-x^3=64\)

giúp mình nha

\(\left(m-1\right)\left(-2\right)+2m-3=0\Leftrightarrow2-2m+2m-3=0\Leftrightarrow-1=0\left(\text{voli}\right)\)

(m−1)(−2)+2m−3=0⇔2−2m+2m−3=0⇔−1=0(voli)

hình như sai đề câu b vs d bn ơi

x là nhân ak

khó quá

xin lỗi nhé mik ko làm đc

3^x ???

=>3^x*8=24

=>3^x=3

=>x=1

1.

A = | x | + 3

vì | x | \(\ge\)0 nên | x | + 3 \(\ge\)3

\(\Rightarrow\)GTNN của A = 3 khi | x | = 0 hay x = 0

tương tự

2.

M = 5 - | x |

vì | x | \(\ge\)0 nên 5 - | x  | \(\le\)5

\(\Rightarrow\)GTLN của M = 5 khi | x | = 0 hay x = 0