a: 12:x=144

=>\(x=\dfrac{12}{144}\)

=>\(x=\dfrac{1}{12}\)

b: \(x-17=\left(-2\right)\cdot27\)

=>\(x-17=-54\)

=>\(x=-54+17=-37\)

c: \(3x-125=145\)

=>\(3x=125+145=270\)

=>\(x=\dfrac{270}{3}=90\)

\(a,12:x=144\\ x=\dfrac{12}{144}=\dfrac{1}{12}\\ ---\\ b,x-17=\left(-2\right).27\\ x-17=-54\\ x=-54+17=-37\\ ----\\ 3x-125=145\\ 3x=145+125=270\\ x=\dfrac{270}{3}=90\)

a: 12:x=144

x=\(\dfrac{1}{12}\)

b: 

c: 

x=90

a: \(4x^2-9=0\)

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

=>x=3/2 hoặc x=-3/2

b: \(5x^2+20=0\)

nên \(x^2+4=0\)(vô lý)

c: \(2x^2-2+\sqrt{3}=0\)

\(\Leftrightarrow2x^2=2-\sqrt{3}\)

\(\Leftrightarrow x^2=\dfrac{4-2\sqrt{3}}{4}\)

hay \(x\in\left\{\dfrac{\sqrt{3}-1}{2};\dfrac{-\sqrt{3}+1}{2}\right\}\)

a) x² - 145 = -64

x²=-64+145

x²=81

x²=9²

=>x=+9

Vậy x=+9

b) (3x - 4)(5x+15)=0

.\(\Rightarrow\orbr{\begin{cases}3x-4=0\\5x+15=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=4\Rightarrow x=\frac{4}{3}\\5x=-15\Rightarrow x=-15:5=-3\end{cases}}\)

Vậy..............................................

a) x² - 145 = -64

x² = -64 +145

x²= 81

=> x²= 9²

=> x = 9

Vậy...

\(a,x^2-145=-64\)

\(x^2=81\)

\(x^2=9^2\)

\(\Rightarrow\orbr{\begin{cases}x=9\\x=-9\end{cases}}\)

a) Tính bằng cách thuận tiện nhất:

134 x 4 x 5

= 134 x (4 x 5)

= 134 x 20 = 1680

5 x 36 x 2

= 36x (5 x 2)

= 36 x 10 = 360

42 x 2 x 7 x5

= (42 x 7) x (2 x5)

= 294 x 10 = 2940

b) 137 x 3 + 137 x 97

= 137 x (3 + 97)

= 137 x 100 = 13700

94 x 12 + 94 x 88

= 94 x (12 + 88)

= 94 x 100 = 9400

428 x 12 - 428 x 2

= 428 x (12 - 2) = 4280

537 x 39 - 537 x 19

= 537 x (39 - 19)

= 537 x 20 = 10740

a) \(\left[\left(x+32\right)-17\right].2=42\)

\(\left[\left(x+32\right)-17\right]=42:2\)

\(\left[\left(x+32\right)-17\right]=21\)

\(\left(x+32\right)=21+17\)

\(\left(x+32\right)=38\)

\(x=38-32\)

\(x=6\)

b) \(125+\left(145-x\right)=175\)

\(\left(145-x\right)=175-125\)

\(\left(145-x\right)=50\)

\(x=145-50\)

\(x=95\)

a) Ta có: \(x^2-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x-1=0\)hay x=1

Vậy: S={1}

c) Ta có: \(x+x^4=0\)

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

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

mà \(x^2-x+1>0\forall x\)

nên x(x+1)=0

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

Vậy: S={0;-1}

a: \(x^3-4x^2-x+4=0\)

=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)

=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(x^2-1\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

=>\(\left(x+1\right)^3=0\)

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)

=>\(\left(x+3\right)\left(x^2-4\right)=0\)

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

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

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)

=>\(\left(x-2\right)\left(x-2-4\right)=0\)

=>(x-2)(x-6)=0

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

a) 12 - 6 = 6

b) 23 - (-35) = 58

c) = 109

Bài 2: a) = 1

b) -3

c) = 8 

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