Bài 1:

a) x³ - 3x² + 3x - 1 + 2(x² - x)

= (x - 1)³ + 2x(x - 1)

= (x - 1)[(x - 1)² + 2x]

= (x - 1)(x² - 2x + 1 + 2x)

= (x - 1)(x² + 1)

b) 36 - 4a² + 20ab - 25b²

= 36 - (2a - 5b)²

= (6 - 2a + 5b)(6 + 2a - 5b)

c) 5a³ - 10a²b + 5ab² - 10a + 10b

= 5(a³ - 2a²b + ab² - 2a + 2b)

= 5[a(a² - 2ab + b²) - 2(a - b)]

= 5[a(a - b)² - 2(a - b)]

= 5(a - b)(a² - ab - 2)

\(10a=15b=6c\)

\(\Rightarrow\frac{10a}{1}=\frac{5b}{\frac{1}{3}}=\frac{c}{\frac{1}{6}}\)

Áp dụng t/c dãy tỉ số bằng nhau, ta có:

\(\frac{10a}{1}=\frac{5b}{\frac{1}{3}}=\frac{c}{\frac{1}{6}}=\frac{10a-5b+c}{1-\frac{1}{3}+\frac{1}{6}}=\frac{25}{\frac{5}{6}}=30\)

\(\Rightarrow\hept{\begin{cases}a=30:10=3\\b=10:5=2\\c=30:6=5\end{cases}}\)

Vậy a = 3, b = 2, c = 5

#)Giải :

Ta có : \(10a=15b\Rightarrow\frac{a}{15}=\frac{b}{10}\Rightarrow\frac{a}{90}=\frac{b}{60}\)

            \(15b=6c\Rightarrow\frac{b}{6}=\frac{c}{15}\Rightarrow\frac{b}{60}=\frac{c}{150}\)

\(\Rightarrow\frac{a}{90}=\frac{b}{60}=\frac{c}{150}\)

Áp dụng tính chất dãy tỉ số bằng nhau :

\(\frac{a}{90}=\frac{b}{60}=\frac{c}{150}=\frac{10a-5b+c}{900-300+150}=\frac{25}{750}=\frac{1}{30}\)

\(\Rightarrow\frac{a}{90}=\frac{1}{30}\Rightarrow a=3\)

\(\Rightarrow\frac{b}{60}=\frac{1}{30}\Rightarrow b=2\)

\(\Rightarrow\frac{c}{150}=\frac{1}{30}\Rightarrow c=5\)

Phối hợp cả 3 phương phép để phân tích các đa thức sau thành phân tử:

a) 36 - 4a² + 20ab - 25b²

= 36 - (4a² - 20ab + 25b²)

= 6² - (2a - 5b)²

= (6 - 2a + 5b)(6 + 2a - 5b)

b) a³ + 3a² + 3a + 1 - 27b³

= (a + 1)³ - (3b)³

= (a + 1 - 3b)[(a + 1)² + 3b(a + 1) + 9b²]

= (a + 1 - 3b)(a² + 2a + 1 + 3ab + 3b + 9b²)

c) x² + 2xy + y² - xz - yz

= (x + y)² - z(x + y)

= (x + y)(x + y - z)

d) 5a³ - 10a²b + 5ab² - 10a + 10b

= 5(a³ - 2a²b + ab² - 2a + 2b)

= 5[a(a² - 2ab + b²) - 2(a - b)]

= 5[a(a - b)² - 2(a - b)]

= 5(a - b)(a² - ab - 2)

a) \(3xy^2-12xy+12x\)

\(=3x\left(y-4y+4\right)\)

b) \(3x^3y-6x^2y-3xy^3-6axy^2-3a^2xy+3xy\)

\(=3xy\left(x^2-2x-y^2-2ay-a^2+1\right)\)

\(=3xy\left[\left(x^2-2\cdot x\cdot1+1^2\right)-\left(y^2+2\cdot y\cdot a+a^2\right)\right]\)

\(=3xy\left[\left(x-1\right)^2-\left(y+a\right)^2\right]\)

\(=3xy\left(x-1-y-a\right)\left(x-1+y+a\right)\)

c) \(36-4a^2+20ab-25b^2\)

\(=6^2-\left[\left(2a\right)^2-2\cdot2a\cdot5b+\left(5b\right)^2\right]\)

\(=6^2-\left(2a-5b\right)^2\)

\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)

d) \(5a^3-10a^2b+5ab^2-10a+10b\)

\(=5a\left(a^2-2ab+b^2\right)-10\left(a-b\right)\)

\(=5a\left(a-b\right)^2-10\left(a-b\right)\)

\(=\left(a-b\right)\left[5a\left(a-b\right)-10\right]\)

\(=5\left(a-b\right)\left[a\left(a-b\right)-2\right]\)

\(=5\left(a-b\right)\left(a^2-ab-2\right)\)

a. 3xy²-12xy+12x

= 3x(y²-4y+4)

= 3x(y-2)²

b. 3x³y-6x²y-3xy³-6axy²-3a²xy+3xy

= 3xy( x²-2x-y²-2ay-a²+1)

= 3xy ((x²-2x+1)-(a²-2ay-y²))

=3xy((x-1)²-(a-y)²)

= 3xy((x-1+a-y)(x-1-(a-y))

=3xy(x-1+a-y)(x-1-a+y)

d. =( 5a(a²-2ab+b²))-(10(a+b))

= 5a(a-b)²-10(a-b)

=5a(a-b)(a-b)-10(a-b)

=(a-b)(5a(a-b)-10)

Hình như mik làm sai hết rồi

Lời giải:
a. Áp dụng tính chất dãy tỉ số bằng nhau:

$\frac{A}{3}=\frac{B}{4}=\frac{C}{5}=\frac{A+B+C}{3+4+5}=\frac{180^0}{12}=15^0$

$\Rightarrow A=3.15^0=45^0; B=4.15^0=60^0; C=5.15^0=75^0$

b. Áp dụng TCDTSBN:

$A=2B=6C$

$= A=\frac{B}{\frac{1}{2}}=\frac{C}{\frac{1}{6}}$

$=\frac{A+B+C}{1+\frac{1}{2}+\frac{1}{6}}=\frac{180^0}{\frac{5}{3}}=108^0$

$\Rightarrow A=108^0; B=108^0.\frac{1}{2}=54^0; C=108^0.\frac{1}{6}=18^0$

\(a,36-4a^2+20ab-25b^2\)

\(=6^2-\left(2a-5b\right)^2=\left(6-2a+5b\right)\left(6+2a-5b\right)\)\(b,x^2+2xy+y^2-xz-yz\)

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

\(=\left(x+y\right)\left(x+y-z\right)\)

\(d,5a^2-10a^2b+5ab^2-10a+10b\)

\(=5a^2-5a^2b-5a^2b+5ab^2-10a+10b\)

\(=5a\left(a-b\right)-5ab\left(a-b\right)-10\left(a-b\right)\)

\(=\left(a-b\right)\left(5a-5ab-10\right)\)

\(A=\dfrac{ab+10b+25}{ab+5a+5b+25}+\dfrac{bc+10c+25}{bc+5b+5c+25}+\dfrac{ca+10a+25}{ac+5a+5c+25}\)

\(=\dfrac{\left(ab+5b\right)+\left(5b+25\right)}{\left(ab+5a\right)+\left(5b+25\right)}+\dfrac{\left(bc+5c\right)+\left(5c+25\right)}{\left(bc+5b\right)+\left(5c+25\right)}+\dfrac{\left(ca+5a\right)+\left(5a+25\right)}{\left(ac+5a\right)+\left(5c+25\right)}\)

\(=\dfrac{b\left(a+5\right)+5\left(b+5\right)}{a\left(b+5\right)+5\left(b+5\right)}+\dfrac{c\left(b+5\right)+5\left(c+5\right)}{b\left(c+5\right)+5\left(c+5\right)}+\dfrac{a\left(c+5\right)+5\left(a+5\right)}{a\left(c+5\right)+5\left(c+5\right)}\)

\(=\dfrac{b\left(a+5\right)+5\left(b+5\right)}{\left(a+5\right)\left(b+5\right)}+\dfrac{c\left(b+5\right)+5\left(c+5\right)}{\left(b+5\right)\left(c+5\right)}+\dfrac{a\left(c+5\right)+5\left(a+5\right)}{\left(a+5\right)\left(c+5\right)}\)

\(=\dfrac{b}{b+5}+\dfrac{5}{a+5}+\dfrac{c}{c+5}+\dfrac{5}{b+5}+\dfrac{a}{a+5}+\dfrac{5}{c+5}\)

\(=\left(\dfrac{b}{b+5}+\dfrac{5}{b+5}\right)+\left(\dfrac{a}{a+5}+\dfrac{5}{a+5}\right)+\left(\dfrac{c}{c+5}+\dfrac{5}{c+5}\right)\)

\(=1+1+1=3\) (\(a;b;c\ne-5\))

\(A=\dfrac{ab+5b+5b+25}{a\left(b+5\right)+5\left(b+5\right)}+\dfrac{bc+5c+5c+25}{b\left(c+5\right)+5\left(c+5\right)}+\dfrac{ca+5a+5a+25}{a\left(c+5\right)+5\left(c+5\right)}\)

\(A=\dfrac{b\left(a+5\right)+5\left(b+5\right)}{\left(a+5\right)\left(b+5\right)}+\dfrac{c\left(b+5\right)+5\left(c+5\right)}{\left(b+5\right)\left(c+5\right)}+\dfrac{a\left(c+5\right)+5\left(a+5\right)}{\left(a+5\right)\left(c+5\right)}\)

\(A=\dfrac{b}{b+5}+\dfrac{5}{a+5}+\dfrac{c}{c+5}+\dfrac{5}{b+5}+\dfrac{a}{a+5}+\dfrac{5}{c+5}\)

\(A=\dfrac{a+5}{a+5}+\dfrac{b+5}{b+5}+\dfrac{c+5}{c+5}=1+1+1=3\)

\(a,3.5^{x+1}-100=-25\\ 3.5^{x+1}=-25+100\\ 3.5^{x+1}=75\\ 5^{x+1}=75:3\\ 5^{x+1}=25\\ 2^{x+1}=5^2\\ x+1=2\\ x=2-1\\ x=1\)

\(b,4x-26+2x=28\\ 4x+2x-26\\ 6x-26=28\\ 6x=28+26\\ 6x=54\\ x=54:6\\ x=9\)

giups mik, mik cần gấp lắm

\(\dfrac{a+b}{3}=\dfrac{b+c}{4}=\dfrac{c+a}{5}\\ \Rightarrow\left\{{}\begin{matrix}4a+4b=3b+3c\\5a+5b=3c+3a\\5b+5c=4c+4a\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4a+b-3c=0\\4a-5b-c=0\\2a+5b-3c=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}4a+b-3c=0\\4a=5b+c\\3c=2a+5b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}5b+c+b-3c=0\\4a+b-2a-5b=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}6b=2c\\2a=4b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}c=3b\\a=2b\end{matrix}\right.\\ \Rightarrow M=20b+b-21b+2021=2021\)