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#)Giải :
c) ( a + b )3 = (a+b)(a+b)(a+b)
= a(a+b)(a+b) +b(a+b)(a+b)
= (a2+ab)(a+b)+(ab+b2)(a+b)
= (a3+a2b+a2b+ab2)+(a2b+ab2+ab2+b2)
= a3+a2b+a2b+ab2+a2b+ab2+ab2+b2
= a3+a2b+a2b+a2b+ab2+ab2+ab2+b2
= a3+3a2b+3ab2+b2
Vậy : (a+b)3= a3+ 3a2b + 3ab2 + b2 ( dpcm )
#~Will~be~Pens~#
a) \(\left(a+b\right)^2=\left(a+b\right)\left(a+b\right)\)
\(=a\left(a+b\right)+b\left(a+b\right)\)
\(=a^2+ab+ab+b^2\)
\(=a^2+2ab+b^2\)
Vậy \(\left(a+b\right)^2=a^2+2ab+b^2\)
\(\text{Câu 1: }\)
\(a,\left|x-3\right|\ge0\left(\forall x\in N\right)\)
\(\Rightarrow\left|x-3\right|+2020\ge2020\left(\forall x\in N\right)\)
\(\text{Dấu}"="\text{xảy ra}\Leftrightarrow\left|x-3\right|+2020=2020\)
\(\Leftrightarrow\left|x-3\right|=2020-2020\)
\(\Leftrightarrow\left|x-3\right|=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=0+3\)
\(\Leftrightarrow x=3\) \(\text{Vậy }x=3\text{ để H có GTNN}\)
\(b,\left(x-1\right)^2\ge0\left(\forall x\in N\right)\)
\(\Rightarrow\left(x-1\right)^2+2021\ge2021\left(\forall x\in N\right)\)
\(\text{Dấu}"="\text{xảy ra}\Leftrightarrow\left(x-1\right)^2+2021=2021\)
\(\Leftrightarrow\left(x-1\right)^2=2021-2021\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=0+1\)
\(\Leftrightarrow x=1\) \(\text{Vậy }x=1\text{ để B có GTNN}\)
\(\text{Câu 2:}\)
\(\frac{3a^2-b^2}{a^2+b^2}=\frac{3}{4}\)
\(\Rightarrow\left(3a^2-b^2\right).4=\left(a^2+b^2\right).3\)
\(\Rightarrow12a^2-4b^2=3a^2+3b^2\)
\(\Rightarrow12a^2-3a^2=3b^2+4b^2\left(\text{quy tắc chuyển vế}\right)\)
\(\Rightarrow a^2.\left(12-3\right)=b^2.\left(3+4\right)\)
\(\Rightarrow a^2.9=b^2.7\)
\(\Rightarrow\frac{a^2}{b^2}=\frac{7}{9}\left(\text{tính chất của tỉ lệ thức}\right)\)
\(\text{Câu 3:}\)
\(ab=c^2;\frac{a^2+c^2}{b^2+c^2}\left(1\right)\)
\(\text{Thay }c^2=ab\text{ vào }\left(1\right)\)
\(\Rightarrow\frac{a^2+ab}{b^2+ab}=\frac{a.\left(a+b\right)}{b.\left(a+b\right)}=\frac{a}{b}\left(2\right)\)
\(\text{Từ (1) và (2)}\Rightarrow\frac{a^2+c^2}{b^2+c^2}=\frac{a}{b}\left(đpcm\right)\)
\(\text{Câu 4: }\)
\(A=\frac{a-b+c}{a+2b-c}\)
\(\frac{a}{2}=\frac{b}{5}\Rightarrow a=\frac{2}{5}.b;\frac{c}{7}=\frac{b}{5}\Rightarrow c=\frac{7}{5}.b\)
\(\text{Thay }a=\frac{2}{5}.b;c=\frac{7}{5}.b\text{ vào }A\)
\(\Rightarrow A=\frac{\frac{2}{5}.b-b+\frac{7}{5}.b}{\frac{2}{5}.b+2b-\frac{7}{5}.b}=\frac{b.\left(\frac{2}{5}-1+\frac{7}{.5}\right)}{b.\left(\frac{2}{5}+2-\frac{7}{5}\right)}=\frac{\frac{2}{5}-\frac{5}{5}+\frac{7}{5}}{\frac{2}{5}+\frac{10}{5}-\frac{7}{5}}=\frac{\frac{2-5+7}{5}}{\frac{2+10-7}{5}}=\frac{4}{5}:1=\frac{4}{5}\)
\(\text{Vậy }A=\frac{4}{5}\)
Bài làm
a) 2a²x³ - ax³ - a⁴ - x³a² - ax³ - 2x⁴
= 2a²x³ - ax³ - a⁴ - a²x³ - ax³ - 2x⁴
= ( 2a²x³ - a²x³ ) - ( ax³ + ax³ ) - a⁴ - 2ax⁴
= a²x³ - 2ax³ - a⁴ - 2ax⁴
b) 3xx⁴ + 4xx³ - 5x²x³ - 5x²x²
= 3x⁵ + 4x⁴ - 5x⁵ - 5x⁴
= ( 3x⁵ - 5x⁵ ) + ( 4x⁴ - 5x⁴ )
= -2x⁵ - x⁴
c) 3a - 4b² - 0,8b . 4b² - 2ab . 3b + b . 3b² - 1
= 3a - 4b² - 3,2b³ - 6ab² + 3b³ - 1
= 3a - 4b² - 0,2b³ - 6ab² - 1
d) 5x.2y² - 5x.3xy - x²y + 6xy²
= 10xy² - 15x²y - x²y + 6xy²
= ( 10xy² + 6xy² ) - ( 15x²y + x²y )
= 16xy² - 16x²y
1) \(\left(A+B\right)^2=\left(A+B\right)\left(A+B\right)=A\left(A+B\right)+B\left(A+B\right)\)
\(=A^2+AB+AB+B^2=A^2+2AB+B^2\)
2) \(\left(A-B\right)^2=\left(A-B\right)\left(A-B\right)=A\left(A-B\right)-B\left(A-B\right)\)
\(=A^2-AB-AB+B^2=A^2-2AB+B^2\)
3) \(A^2-B^2=A^2-AB-B^2+AB\)
\(=A\left(A-B\right)+B\left(A-B\right)=\left(A-B\right)\left(A+B\right)\)
p/s: mấy cái kia tương tự