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}

bài2

a, x-15=-63-4

=>x-15=-67

=>x=-52

b, -x+3=11

=>x=-11+3

=>x=-8

c,\(|\)x+2\(|\)-4=7

=>\(|\)x+2\(|\)=11

=>\(\left\{{}\begin{matrix}x+2=11\\x+2=-11\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=9\\x=13\end{matrix}\right.\)

bài3

ta có:\(\left|y\right|\)=8

=>\(\left[{}\begin{matrix}y=8\\y=-8\end{matrix}\right.\)

TH1 x=5,y=8

=>x-y=5-8=-3

y-x=8-5=3

TH2x=5 ,y=-8

x-y=5--8=13

y-x=-8-5=-13

Baif:

a) x-15=-63-4

x-15=-67

x=-67+15

x=-52

b)-x+3=11

-x=11-3

-x=8

=> x=8

c)\(\left|x+2\right|-4=7\)

\(\left|x+2\right|\)=7+4=11

=> x+2=11 hoặc x+2=-11

x=11-2=9 hoặc x=-11-2=-13

Bài 3:

TH1: Nếu x=5 và y=8

thì x-y=5-8=-3

y-x=8-5=3

TH

: Nếu x=5 và y=-8

thì x-y=5-(-8)=13

y-x=(-8)-5=-13

a) x-15 = -63-4

x = -63-4+15

x = -52

b) -x + 3 = 11

-x= 11-3

-x= 8

x=8

c) | x+2 | -4 = 7

|x+2| = 11

⇒ \(\left\{{}\begin{matrix}x+2=11\\x+2=-11\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)

Bài 1:

Từ P(x) = 3x²+8x-4 = -4

=> 3x²+8x = 0

x(3x+8) = 0

=> x = 0 3x+8 = 0

=> x = 0 3x = 8

=> x = 8/3

Bài 2 :

Ta có x = -1 là nghiệm của đa thức f(x) = 2x²-x+m

=> f(-1) = 2(-1)²-(-1)+m = 0

=> 2+1+m = 0

=> 3+m = 0

m = 0-3

m = -3

d, \(\left(3x-2^4\right).7^3=2.7^4\)

\(\Rightarrow3x-2^4=2.7^4:7^3\)

\(\Rightarrow3x-16=2.7\\ \Rightarrow3x=14+16\\ \Rightarrow3x=30\Rightarrow x=10\)

Vậy.....

e, \(x-\left[42+\left(-28\right)\right]=-8\)

\(\Rightarrow x-14=-8\\ \Rightarrow x=6\)

Vậy.....

g, \(x-7=-5\)

\(\Rightarrow x=-5+7\Rightarrow x=2\)

Vậy.....

h, \(15-5\left(x+4\right)=-12-3\)

\(\Rightarrow15-5x-20=-15\)

\(\Rightarrow-5x=-15-15+20\)

\(\Rightarrow-5x=-10\Rightarrow x=2\)

Vậy.....

Chúc bạn học tốt!!!

d/ \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)

\(\Rightarrow3x-16=\dfrac{2\cdot7^4}{7^3}=14\)

\(\Rightarrow3x=14+16=30\)

\(\Rightarrow x=\dfrac{30}{3}=10\)

e/ Đễ ==> tự lm thì tốt hơn nhé

g/ Đễ ==> tự lm thì tốt hơn nhé

h/ \(15-5\left(x+4\right)=-12-3\)

\(\Rightarrow15-5x-20=-15\)

\(\Rightarrow-5x=-15+20-15=-10\)

\(\Rightarrow x=\dfrac{-10}{-5}=2\)

i/ \(\left(7-x\right)-\left(25+7\right)=-25\)

\(\Rightarrow7-x-25-7=-25\)

\(\Rightarrow-x=-25-7+7+25\)

\(\Rightarrow-x=0\Rightarrow x=0\)

k/ \(\left|x+2\right|=0\Rightarrow x+2=0\Rightarrow x=-2\)

l/ \(\left|x-3\right|=7-\left(-2\right)\)

\(\Rightarrow\left|x-3\right|=9\)

\(\Rightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\)

m/ \(\left|x-5\right|=\left|-7\right|\Rightarrow\left|x-5\right|=7\)

\(\Rightarrow\left[{}\begin{matrix}x-5=7\\x-5=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=12\\x=-2\end{matrix}\right.\)

i, \(\left(7-x\right)-\left(25+7\right)=-25\)

\(\Rightarrow7-x-25-7=-25\)

\(\Rightarrow-x=-25-7+25+7\)

\(\Rightarrow x=0\)

Vậy.....

k, \(\left|x+2\right|=0\)

\(\Rightarrow x+2=0\Rightarrow x=-2\)

Vậy.....

l, \(\left|x-3\right|=-7-\left(-2\right)\)

\(\Rightarrow\left|x-3\right|=-5\)

Với mọi giá trị của \(x\in Z\) ta có:

\(\left|x-3\right|\ge0\) mà \(-5< 0\) nên không tìm được giá trị nào của x thoả mãn \(\left|x-3\right|=-5\).

Vậy \(x\in\varnothing\)

m, \(\left|x-5\right|=\left|-7\right|\)

\(\Rightarrow\left|x-5\right|=7\\ \Rightarrow\left\{{}\begin{matrix}x-5=-7\\x-5=7\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-2\\x=12\end{matrix}\right.\)

Vậy...,..

Chúc bạn học tốt!!!

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)

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

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

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

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

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

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

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

Bài 3:

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

⇒ \(\left|1-2x\right|+x=0-2\)

⇒ \(\left|1-2x\right|+x=-2\)

⇒ \(\left|1-2x\right|=-2-x\)

⇒ \(\left[{}\begin{matrix}1-2x=-2-x\\1-2x=2+x\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}1+2=-x+2x\\1-2=x+2x\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}3=1x\\-1=3x\end{matrix}\right.\)

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

Vậy \(x\in\left\{3;-\frac{1}{3}\right\}.\)

Bài 4:

\(\left|5x-3\right|=\left|7-x\right|\)

⇒ \(\left[{}\begin{matrix}5x-3=7-x\\5x-3=x-7\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}5x+x=7+3\\5x-x=\left(-7\right)+3\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}6x=10\\4x=-4\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}x=10:6\\x=\left(-4\right):4\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\frac{5}{3}\\x=-1\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{5}{3};-1\right\}.\)

Chúc bạn học tốt!

hình như sai đề câu b vs d bn ơi

x là nhân ak

khó quá

xin lỗi nhé mik ko làm đc

Mik ko biết làm phần a nha

b) Ta có: (x+3)⋮(x−4)(x+3)⋮(x−4)

⇒(x+3)−(x−4)⋮(x−4)⇒(x+3)−(x−4)⋮(x−4)

⇒(x+3−x+4)⋮(x−4)⇒(x+3−x+4)⋮(x−4)

⇒7⋮(x−4)⇒7⋮(x−4)

⇒(x−4)∈Ư(7)⇒(x−4)∈Ư(7)

⇒x−4∈{−1;1;7;−7}⇒x−4∈{−1;1;7;−7}

⇒x∈{3;5;11;−3}⇒x∈{3;5;11;−3}

Vậy: x∈{3;5;11;−3}x∈{3;5;11;−3}

c)

Ta có: x-5 là bội của 7-x

⇒(x−5)⋮(7−x)⇒(x−5)⋮(7−x)

⇒(x−5)+(7−x)⋮(7−x)⇒(x−5)+(7−x)⋮(7−x)

⇒(x−5+7−x)⋮(7−x)⇒(x−5+7−x)⋮(7−x)

⇒2⋮(7−x)⇒2⋮(7−x)

⇒(7−x)∈Ư(2)⇒(7−x)∈Ư(2)

⇒(7−x)∈{−1;1;2;−2}⇒(7−x)∈{−1;1;2;−2}

⇒x∈{8;6;5;9}⇒x∈{8;6;5;9}

Vậy: x∈{8;6;5;9}