a) y x 4,9 : 10 = 12

    y x 49         = 12

    y                 = 12 : 49

    y                 =    tự tính

\(y\times4,9+y\div10=1,2\)

\(y\times4,9+y\times\dfrac{1}{10}=1,2\)

\(y\times4,9+y\times0,1=1,2\)

\(y\times\left(4,9+0,1\right)=1,2\)

\(y\times5=1,2\)

\(y=1,2\div5\)

\(y=0,24\)

a, x=15
b, x=11
c, x=7
d, x=36

Bài 6:

a)(x-15).15=0

⇔x-15=0

⇔x=15

b)32(x-10)=32

⇔x-10=1

⇔x=11

c)(x-5)(x-7)=0

⇔x-5=0 hay x-7=0

⇔x=5 hay x=7

d)(x-35)35=35

⇔x-35=1

⇔x=36

a, x = 15
b, x = 11
c, x = 7 hoặc x = 5
d, x = 36

hơi khó đấy bạn

Bài 6:

a)(x-15).15=0

⇔x-15=0

⇔x=15

b)32(x-10)=32

⇔x-10=1

⇔x=11

c)(x-5)(x-7)=0

⇔x-5=0 hay x-7=0

⇔x=5 hay x=7

d)(x-35)35=35

⇔x-35=1

⇔x=36

a: =>7(x-5)>0

=>x-5>0

=>x>5

b: =>x-1 thuộc {1;-1;11;-11}

=>x thuộc {2;0;12;-10}

c: =>x+1+7 chia hết cho x+1

=>x+1 thuộc {1;-1;7;-7}

=>x thuộc {0;-2;6;-8}

d: =>(x+2)(x-5)<0

=>-2<x<5

a:(- 7) . ( 5 – x) < 0

=>7(x-5)>0

=>x-5>0

=>x>5

b:11 ⁝ x – 1

=>x-1 thuộc {1;-1;11;-11}

=>x thuộc {2;0;12;-10}

c: x + 8 ⁝ x + 1

=>x+1+7 chia hết cho x+1

=>x+1 thuộc {1;-1;7;-7}

=>x thuộc {0;-2;6;-8}

d: (x + 2) . (5 – x) > 0

=>(x+2)(x-5)<0

=>-2<x<5

a) Để x > 3.1 và < 3.6 => x = 3.2; 3.3; 3.4; 3.5; 3.56

b) Để x > 2.1 và < 2.2 => x = 2.11; 2.12; 2.13; 2.15; 2.18

Tìm 5 giá trị x sao cho

a) 3,1 < x < 3,6

b) 2,1 < x < 2,2

a)  x = 3.2; 3.3; 3.4; 3.5; 3.56

b)  x = 2.11; 2.12; 2.13; 2.15; 2.18

Câu 2:

\(\dfrac{a+b}{6}=\dfrac{b+c}{7}=\dfrac{c+a}{8}=\dfrac{2\left(a+b+c\right)}{6+7+8}=\dfrac{28}{21}=\dfrac{4}{3}\\ \Rightarrow\left\{{}\begin{matrix}a+b=\dfrac{4}{3}\cdot6=8\\b+c=\dfrac{4}{3}\cdot7=\dfrac{28}{3}\\c+a=\dfrac{4}{3}\cdot8=\dfrac{32}{3}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a=14-\dfrac{28}{3}=\dfrac{14}{3}\\b=14-\dfrac{32}{3}=\dfrac{10}{3}\\c=14-8=6\end{matrix}\right.\)

Vậy chọn C

\(a,=3x-9-4x+12=-x+3=0\)

\(\Leftrightarrow x=3\)

Vậy ..

\(b,=\left(x+2\right)\left(x+2-x+2\right)=4\left(x+2\right)=0\)

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)

Vậy ..

\(c,=x^3-3x^2+3x-1=\left(x-1\right)^3=0\)

\(\Leftrightarrow x=1\)

Vậy ..

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

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

Vậy ..

\(e,=\left(2x-3-5\right)\left(2x-3+5\right)=\left(2x-8\right)\left(2x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{2}=4\\x=-\dfrac{2}{2}=-1\end{matrix}\right.\)

Vậy ...

a) Ta có: 3(x-3)-4x+12=0

\(\Leftrightarrow3\left(x-3\right)-4\left(x-3\right)=0\)

\(\Leftrightarrow x-3=0\)

hay x=3

Vậy: S={3}

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

\(\Leftrightarrow x^2+4x+4-x^2+4=0\)

\(\Leftrightarrow4x=-8\)

hay x=-2

Vậy: S={-2}

c) Ta có: \(x^3+3x=3x^2+1\)

\(\Leftrightarrow x^3-3x^2+3x-1=0\)

\(\Leftrightarrow x-1=0\)

hay x=1

Vậy: S={1}

d) Ta có: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)

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

Vậy: S={0;2;-2}

a) 3.(x-3)-4x+12=0

=> 3x - 9 - 4x + 12 = 0

=> -x + 3 = 0

=> x = 3

b) (x+2)^2-(x+2).(x-2) =0

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

\(\Rightarrow x^2+4x+4-x^2+4=0\)

=> 4x + 8 = 0

=> x = -2

c) x^3+3x=3x^2+1

\(\Rightarrow x^3+3x-3x^2-1=0\)

\(\Rightarrow\left(x-1\right)^3=0\)

=> x = 1

d) \(\dfrac{2}{3}x\left(x^2-4\right)=0\)

\(\Rightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)

=> x = 0 hoặc x = 2 hoặc x = -2

e) \(\left(2x-3\right)^2-5^2=0\)

\(\Rightarrow\left(2x-8\right)\left(2x+2\right)=0\)

=> x = 4 hoăc x = -1

\(a,x\in\left\{-5;-4;-3;-2;-1\right\}\\ b,x\in\left\{-3;-2;-1;...;5;6\right\}\\ c,x\in\left\{-4;-3;...;3;4\right\}\\ d,x\in\left\{-3;-2;-1;0;1;2\right\}\)