K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

24 tháng 6 2021

tách nhỏ ra đi ak

Bài 5:

1) \(\left(5+7\right)\left(7-5\right)=7^2-5^2\)

2) \(\left(x+y\right)\left(y-x\right)=y^2-x^2\)

3) \(\left(x-y\right)\left(-x-y\right)=-\left(x+y\right)\left(x-y\right)=-\left(x^2-y^2\right)=y^2-x^2\)

6) \(\left(2+3x^2\right)\left(3x^2-2\right)=9x^4-4\)

7) \(\left(\dfrac{1}{2}+x\right)\left(-x+\dfrac{1}{2}\right)=\left(\dfrac{1}{2}+x\right)\left(\dfrac{1}{2}-x\right)=\dfrac{1}{4}-x^2\)

8) \(\left(4m-5n\right)\left(5n+4m\right)=\left(4m-5n\right)\left(4m+5n\right)=16m^2-25n^2\)

9) \(\left(7a+1\right)\left(1-7a\right)=\left(1+7a\right)\left(1-7a\right)=1-49a^2\)

10) \(\left(1+9\right)\left(1-9\right)=1-9^2\)

25 tháng 5 2022

a) \(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}\\\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{y}{15}=\dfrac{z}{21}\end{matrix}\right.\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x+y+z}{10+15+21}=\dfrac{92}{46}=2\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{10}=2\Rightarrow x=2.10=20\\\dfrac{y}{15}=2\Rightarrow y=2.15=30\\\dfrac{z}{21}=2\Rightarrow z=2.21=42\end{matrix}\right.\)

6 tháng 9 2017

Bình phương 1 tổng: \(\left(A+B\right)^2=A^2+2AB+B^2\)

\(\left(2+3y\right)^3=2^3+3.2^2.3y+3.2.\left(3y\right)^2+\left(3y\right)^3\)

\(=8+36y+54y^2+27y^3\)

14 tháng 7 2019

a) \(\left(2x+1\right)^2-4\left(x+2\right)^2=12\)

\(\Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)=12\)

\(\Leftrightarrow4x^2+4x+1-4x^2-16x-16-12=0\)

\(\Leftrightarrow-12x-27=0\)

\(\Leftrightarrow x=\frac{-9}{4}\)

b) xem lại đề

c) \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x-3\right)\left(3-x\right)=1\)

\(\Leftrightarrow x^3-27-x\left(x-3\right)^2=1\)

\(\Leftrightarrow x^3-27-x\left(x^2-6x+9\right)-1=0\)

\(\Leftrightarrow x^3-28-x^3+6x^2-9x=0\)

\(\Leftrightarrow6x^2-9x-28=0\)

\(\Leftrightarrow6\left(x^2-\frac{3}{2}x-\frac{14}{3}\right)=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\frac{3}{4}+\frac{9}{16}-\frac{251}{48}=0\)

\(\Leftrightarrow\left(x-\frac{3}{4}\right)^2=\frac{251}{48}=\left(\pm\sqrt{\frac{251}{48}}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{3}{4}=\sqrt{\frac{251}{48}}=\frac{\sqrt{753}}{12}\\x-\frac{3}{4}=-\sqrt{\frac{251}{48}}=\frac{-\sqrt{753}}{12}\end{matrix}\right.\)

\(\Leftrightarrow x=\frac{\pm\sqrt{753}}{12}+\frac{3}{4}=\frac{9\pm\sqrt{753}}{12}\)

d) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-19\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+19=0\)

\(\Leftrightarrow12x+15=0\)

\(\Leftrightarrow x=\frac{-5}{4}\)

14 tháng 7 2019

Theo giả thiết:

\(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Dễ thấy \(VT\ge0\forall a;b;c\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\)\(\Leftrightarrow a=b=c\)(đpcm)

12 tháng 10 2023

5: \(=4b^2-2b+\dfrac{1}{4}-\dfrac{1}{4}+a-a^2\)

\(=\left(2b\right)^2-2\cdot2b\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\left(a^2-a+\dfrac{1}{4}\right)\)

\(=\left(2b-\dfrac{1}{2}\right)^2-\left(a-\dfrac{1}{2}\right)^2\)

\(=\left(2b-\dfrac{1}{2}-a+\dfrac{1}{2}\right)\left(2b-\dfrac{1}{2}+a-\dfrac{1}{2}\right)\)

\(=\left(2b-a\right)\left(2b+a-1\right)\)

6:

\(=b^2-4b+4-9c^2\)

\(=\left(b-2\right)^2-9c^2\)

\(=\left(b-2-3c\right)\left(b-2+3c\right)\)

12 tháng 10 2023

5) \(4b^2-2b+a-a^2\)

\(=\left(2b-a\right)\left(2b+a\right)-\left(2b-a\right)\)

\(=\left(2b-a\right)\left(2b+a-1\right)\)

6) \(b^2-9c^2+4+4b\)

\(=\left(b+2\right)^2-9c^2\)

\(=\left(b+3c+2\right)\left(b-3c+2\right)\)

28 tháng 10 2021

=(3x) - 1 = 9x2 - 1

28 tháng 10 2021

=\(9x^2-1\)