\(4-\frac{13}{6}=\frac{24}{6}-\frac{13}{6}=\frac{24-13}{6}=\frac{11}{6}\)

4 - \(\frac{13}{6}\) = \(\frac{24}{6}\) - \(\frac{13}{6}\) = \(\frac{11}{6}\)

2\(^3\) = 2.2.2 = 4.2 = 8

vì 2x2x2 = 2 mũ 3

\(\frac47+3=\frac47+\frac{21}{7}=\frac{4+21}{7}=\frac{25}{7}\)

4/7 + 3

= 4/7 + 21/7 = 4+21/7 = 25/7

Bài 14:

\(A\left(x\right)+B\left(x\right)=5x^4-6x^3-3x^2-4\)

\(A\left(x\right)-B\left(x\right)=3x^4+7x^2+8x+2\)

Do đó: \(A\left(x\right)+B\left(x\right)+A\left(x\right)-B\left(x\right)=5x^4-6x^3-3x^2-4+3x^4+7x^2+8x+2\)

=>\(2\cdot A\left(x\right)=8x^4-6x^3+4x^2+8x-2\)

=>\(A\left(x\right)=4x^4-3x^3+2x^2+4x-1\)

Ta có: \(A\left(x\right)+B\left(x\right)=5x^4-6x^3-3x^2-4\)

=>\(B\left(x\right)=5x^4-6x^3-3x^2-4-4x^4+3x^3-2x^2-4x-1\)

=>\(B\left(x\right)=x^4-3x^3-5x^2-4x-5\)

Bài 13:

\(f\left(x\right)+g\left(x\right)=6x^4-3x^2-5\)

\(f\left(x\right)-g\left(x\right)=4x^4-6x^3+7x^2+8x-9\)

Do đó: \(f\left(x\right)+g\left(x\right)+f\left(x\right)-g\left(x\right)=6x^4-3x^2-5+4x^4-6x^3+7x^2+8x-9\)

=>\(2\cdot f\left(x\right)=10x^4-6x^3+4x^2+8x-14\)

=>\(f\left(x\right)=5x^4-3x^3+2x^2+4x-7\)

\(f\left(x\right)+g\left(x\right)=6x^4-3x^2-5\)

=>\(g\left(x\right)=6x^4-3x^2-5-5x^4+3x^3-2x^2-4x+7=x^4+3x^3-5x^2-4x+2\)

\(\frac37\) x (\(x\) - \(\frac{14}{9}\)) = - \(\frac{11}{7}\) x (\(x+\frac{14}{11}\))

\(\frac37x\) - \(\frac23\) = - \(\frac{11}{7}x\) - 2

\(\frac37\)\(x\) + \(\frac{11}{7}\)\(x\) = - 2 + \(\frac23\)

2\(x\) = \(-\frac43\)

\(x\) = - \(\frac43:2\)

\(x\) = - \(\frac23\)

Vậy \(x=-\frac23\)

73​ x (\(x\) - \(\frac{14}{9}\)) = - \(\frac{11}{7}\) x (\(x + \frac{14}{11}\))

\(\frac{3}{7} x\) - \(\frac{2}{3}\) = - \(\frac{11}{7} x\) - 2

\(\frac{3}{7}\)\(x\) + \(\frac{11}{7}\)\(x\) = - 2 + \(\frac{2}{3}\)

2\(x\) = \(- \frac{4}{3}\)

\(x\) = - \(\frac{4}{3} : 2\)

\(x\) = - \(\frac{2}{3}\)

Vậy \(x = - \frac{2}{3}\)

a) X = {x ∈ N | 8 < x < 15}
→ X = {9, 10, 11, 12, 13, 14}

b) Y = {x ∈ N | x < 7}
→ Y = {0, 1, 2, 3, 4, 5, 6}

c) Z = {x ∈ N | 13 ≤ x ≤ 20}
→ Z = {13, 14, 15, 16, 17, 18, 19, 20}

d) T = {x ∈ N | 4 ≤ x < 9}
→ T = {4, 5, 6, 7, 8}

a) X={9;10;11;12;13;14}

b) Y={0;1;2;3;4;5;6}

c) Z={13;14;15;16;17;18;19;20}

d) T={4;5;6;7;8}

x =5

y =-6

z=-7

Ta có
:\(x y - y z\)=-30-42

=>\(y \left(\right. x - z \left.\right)\)=-72

=>-12\(y\)=-72

=>\(y\)=6

vì \(x y = - 30\)

\(\Leftrightarrow\) \(x . 6 = - 30\)

\(\Leftrightarrow\)\(x = - 5\)

Vì \(z - x = - 12\)

\(\Leftrightarrow\)\(z - \left(\right. - 5 \left.\right) = - 12\)

\(\Leftrightarrow\)\(z + 5 = - 12\)

\(\Leftrightarrow\)\(z = - 17\)

VẬY \(\left(\right.x;y\left.\right)=\left(\right.6;-5\left.\right)\)

50000cm

 50000 cm

sao nhìn nó lạ lắm ko giống x đâu bn nên bn ghi lại đi để mik nhìn rõ hơn nha :))

\(x\in\left\lbrace45,46\right\rbrace\) nhé

\(1-\frac12=\frac22-\frac12=\frac12\)

1 \(-\frac12\)

\(=\frac22-\frac12=\frac12\) ​
​
Đ/S : \(\frac{1}{2}\)