a) \(\left(x+2\right)^2=25\)

\(\Rightarrow\left(x+2\right)^2=5^2\)

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

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\)

b) x chia hết cho 7 

⇒ \(x\in B\left(7\right)=\left\{0;7;14;21;28;35;42;49;56;...\right\}\)

Mà: \(15< x< 56\)

\(\Rightarrow x\in\left\{21;28;35;42;49\right\}\)

a: =>x+2=5 hoặc x+2=-5

=>x=-7 hoặc x=3

b: x chia hết cho 7

=>x thuộc {0;7;14;21;28;35;42;49;56;63}

mà 15<x<56

nên x thuộc {21;28;35;42;49}

a) Có: (x + 2)² = 25

\(\Leftrightarrow\left[{}\begin{matrix}x+2=5\\x+2=-5\end{matrix}\right.\)    \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\)

b) Có: x ⋮ 7 => x∈ B(7)

mà 15 < x < 56 nên x ∈ { 21 ; 28 ; 35 ; 42 ; 49 }

Vậy...

Đề y/c gì em?

a: =7/35*18/9+3/25

=2/5+3/25

=13/25

b: =1/3+4/3*5/2

=1/3+20/6

=1/3+10/3

=11/3

mk sẽ tích đúng cho

Bài 2: 

a: x=4738

c: x=10254

tk

a: Khi x=25 thì \(A=\dfrac{7}{5+8}=\dfrac{7}{13}\)

b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{x-9}\)

\(=\dfrac{x+5\sqrt{x}-24}{x-9}=\dfrac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{x-9}=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\)

c: P=A*B

\(=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\cdot\dfrac{7}{\sqrt{x}+8}=\dfrac{7}{\sqrt{x}+3}\)

P là số nguyên

=>căn x+3 thuộc Ư(7)

=>căn x+3=7

=>x=16

a, \(\frac{2}{3}:\frac{5}{7}:\frac{x}{9}=\frac{2}{7}:\frac{3}{5}:\frac{10}{9}\Leftrightarrow\frac{2\times7\times9}{3\times5\times x}=\frac{2\times5\times9}{7\times3\times10}\Leftrightarrow\frac{42}{5\times x}=\frac{3}{7}\Leftrightarrow294=15\times x\Leftrightarrow x=19,6\)

Các phần sau tương tự.

Dùng phương pháp nhân chéo và kiểm tra trình độ tính toán, khả năng rút gọn.

Thay kết quả vào x để kiểm tra lại bài.

cho hỏi Xx9/7:3/4=12/15

\(a,\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x-5=0\Leftrightarrow x=5\\ b,\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow2x-1=0\Leftrightarrow x=1\\ c,\Leftrightarrow\left(1-2x\right)^2-\left(3x-2\right)^2=0\\ \Leftrightarrow\left(1-2x-3x+2\right)\left(1-2x+3x-2\right)=0\\ \Leftrightarrow\left(3-5x\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{5}\end{matrix}\right.\\ d,\Leftrightarrow\left(x-2\right)^3=-\left(5-2x\right)^3\\ \Leftrightarrow x-2=-\left(5-2x\right)=2x-5\\ \Leftrightarrow x=3\)

\(a,\left(x-5\right)\left(2x+3\right)=x^2-25\\ \Leftrightarrow a,\left(x-5\right)\left(2x+3\right)-\left(x-5\right)\left(x+5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(2x+3-x+5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-8\end{matrix}\right.\\ b,\dfrac{2x}{3}+\dfrac{2x-1}{6}=\dfrac{x-1}{2}\\ \Leftrightarrow\dfrac{4x}{6}+\dfrac{2x-1}{6}-\dfrac{3\left(x-1\right)}{6}=0\\ \Leftrightarrow4x+2x-1-3x+3=0\\ \Leftrightarrow3x+2=0\\ \Leftrightarrow x=-\dfrac{2}{3}\)

a) \(\left(3x-5\right)\left(3x+5\right)=9x^2-25\Leftrightarrow9x^2+15x-15x-25=9x^2-25\Leftrightarrow9x^2-25=9x^2-25\)(đúng)

b) \(x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)\Leftrightarrow x^3-y^3=x^3+x^2y+xy^2-x^2y-xy^2-y^3\Leftrightarrow x^3-y^3=x^3-y^3\)(đúng)

c) \(x^2+y^2=\left(x+y\right)^2-2xy\Leftrightarrow x^2+y^2=x^2+y^2+2xy-2xy\Leftrightarrow x^2+y^2=x^2+y^2\)(đúng)

a: \(\left(3x-5\right)\left(3x+5\right)\)

\(=9x^2+15x-15x-25\)

\(=9x^2-25\)

b: \(\left(x-y\right)\left(x^2+xy+y^2\right)\)

\(=x^3+x^2y+xy^2-x^2y-xy^2-y^3\)

\(=x^3-y^3\)

c: \(\left(x+y\right)^2-2xy\)

\(=x^2+2xy+y^2-2xy\)

\(=x^2+y^2\)

a: Ta có: \(\left(3x-5\right)\left(3x+5\right)\)

\(=9x^2+15x-15x-25\)

\(=9x^2-25\)

b: Ta có: \(\left(x-y\right)\left(x^2+xy+y^2\right)\)

\(=x^3+x^2y+xy^2-x^2y-xy^2-y^3\)

\(=x^3-y^3\)

c: Ta có: \(\left(x+y\right)^2-2xy\)

\(=x^2+2xy+y^2-2xy\)

\(=x^2+y^2\)