1) |3x-18| = 45

TH1: 3x - 18 = 45

3x = 63

x = 21

TH2: 3x -18 = -45

3x = -27

x = -9

KL:...

phần còn lại b tự làm nha

Ta có \(\left|3x-18\right|=45\Rightarrow\orbr{\begin{cases}3x-18=-45\\3x-18=45\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=-27\\3x=63\end{cases}\Rightarrow\orbr{\begin{cases}x=-9\\x=21\end{cases}}}\)

Vậy ..........

2,\(\left|19+x+95\right|=190\Rightarrow\left|x+114\right|=190\)

\(\Rightarrow\orbr{\begin{cases}x+114=190\\x+114=-190\end{cases}\Rightarrow\orbr{\begin{cases}x=76\\x=-304\end{cases}}}\)

Vậy......

TK MK NHA

1. x(x + 1) - x² + 1 = 0

<=> x(x + 1) - (x² - 1) = 0

<=> x(x + 1) - (x + 1)(x - 1) = 0

<=> (x - x + 1)(x + 1) = 0

<=> x + 1 = 0\

<=> x = -1

2. 4x(x - 2) - 6 + 3x = 0

<=> 4x(x - 2) - (3x - 6) = 0

<=> 4x(x - 2) - 3(x - 2) = 0

<=> (4x - 3)(x - 2) = 0

<=> \(\left[{}\begin{matrix}4x-3=0\\x-2=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=2\end{matrix}\right.\)

3. x(x + 2) - 3(x + 2) = 0

<=> (x - 3)(x + 2) = 0

<=> \(\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)

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

câu 1:

a) 500-(300)-190+(-210)

= 500-300-190-210

= 200 - 210 -190

=-10 - 190

=-200

b) (-3)³ .5+12.(-6)

= -27.5 -72

=-135 - 72

=-207

c) 15.(-19-4)-19.(15-4)

= 15.(-23) - 19.11

=-345 - 209

=-554

câu 2: tìm x thuộc Z

a) 3x-2=3

=> 3x=3/2

=> x=1/2

b) x chia hết cho 5 và -7<x<11

=> x thuộc {-5;0;5;10}

Câu 1: 

a) Ta có: \(500-\left(300\right)-190+\left(-210\right)\)

\(=500-300-190-210\)

\(=\left(500-300\right)-\left(190+210\right)\)

\(=200-400=-200\)

b) Ta có: \(\left(-3\right)^3\cdot5+12\cdot\left(-6\right)\)

\(=\left(-3\right)^3\cdot5-3\cdot4\cdot3\cdot2\)

\(=-5\cdot3^3-3^2\cdot8\)

\(=3^2\cdot\left(-5\cdot3-8\right)\)

\(=9\cdot\left(-15-8\right)=9\cdot\left(-23\right)=-207\)

c) Ta có: \(15\cdot\left(-19-4\right)-19\cdot\left(15-4\right)\)

\(=-15\cdot19-15\cdot4-15\cdot19+19\cdot4\)

\(=-30\cdot19+4\cdot4\)

\(=-2\cdot\left(15\cdot19+2\cdot4\right)\)

\(=-2\cdot\left(285+8\right)=-586\)

1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)

\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)

\(\Leftrightarrow x=2\)

3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)

\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)

\(\Leftrightarrow6x=6\)

hay x=1

1)
\(4x^2-4x+1-4x^2-16x-16=9\)
\(-20x-15=9\)
-20x=24
x=-1,2

3)
(2x+1)²=5²
2x+1=5
2x=4
x=2
 

\(1,\Rightarrow4x^2-4x+1-4x^2-16x-16=9\\ \Rightarrow-20x=23\Rightarrow x=-\dfrac{23}{20}\\ 2,\Rightarrow9x^2-6x+1+2x+6+11-11x^2=15\\ \Rightarrow2x^2+4x-3=0\\ \Rightarrow2\left(x^2+2x+1\right)-5=0\\ \Rightarrow2\left(x+1\right)^2-5=0\\ \Rightarrow\left[\sqrt{2}\left(x+1\right)-\sqrt{5}\right]\left[\sqrt{2}\left(x+1\right)+\sqrt{5}\right]=0\\ \Rightarrow\left[{}\begin{matrix}\sqrt{2}\left(x+1\right)=\sqrt{5}\\\sqrt{2}\left(x+1\right)=-\sqrt{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+1=\sqrt{\dfrac{5}{2}}=\dfrac{\sqrt{10}}{2}\\x+1=-\sqrt{\dfrac{5}{2}}=\dfrac{-\sqrt{10}}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{10}-2}{2}\\x=\dfrac{-\sqrt{10}-2}{2}\end{matrix}\right.\)

\(3,\Rightarrow\left(2x+1\right)^2-25=0\Rightarrow\left(2x+1-5\right)\left(2x+1+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

\(4,\Rightarrow x^3+3x^2+3x+1-x^3-2x^2-2x+1-x^2=15\\ \Rightarrow x+2=15\Rightarrow x=13\)

a)=18-5-7

=13-7

=5

b)=(48-2)+(-21)

46+(-21)

=25

c)=-200.(-2)+20

=400+20=420