tính a+b+c biết
a+b=10
a+c=25
b+c=25
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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:
$A=2B=6C$
$= A=\frac{B}{\frac{1}{2}}=\frac{C}{\frac{1}{6}}$
$=\frac{A+B+C}{1+\frac{1}{2}+\frac{1}{6}}=\frac{180^0}{\frac{5}{3}}=108^0$
$\Rightarrow A=108^0; B=108^0.\frac{1}{2}=54^0; C=108^0.\frac{1}{6}=18^0$
\(a,36-4a^2+20ab-25b^2\)
\(=6^2-\left(2a-5b\right)^2=\left(6-2a+5b\right)\left(6+2a-5b\right)\)\(b,x^2+2xy+y^2-xz-yz\)
\(=\left(x+y\right)^2-z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-z\right)\)
\(d,5a^2-10a^2b+5ab^2-10a+10b\)
\(=5a^2-5a^2b-5a^2b+5ab^2-10a+10b\)
\(=5a\left(a-b\right)-5ab\left(a-b\right)-10\left(a-b\right)\)
\(=\left(a-b\right)\left(5a-5ab-10\right)\)
\(A=\dfrac{ab+10b+25}{ab+5a+5b+25}+\dfrac{bc+10c+25}{bc+5b+5c+25}+\dfrac{ca+10a+25}{ac+5a+5c+25}\)
\(=\dfrac{\left(ab+5b\right)+\left(5b+25\right)}{\left(ab+5a\right)+\left(5b+25\right)}+\dfrac{\left(bc+5c\right)+\left(5c+25\right)}{\left(bc+5b\right)+\left(5c+25\right)}+\dfrac{\left(ca+5a\right)+\left(5a+25\right)}{\left(ac+5a\right)+\left(5c+25\right)}\)
\(=\dfrac{b\left(a+5\right)+5\left(b+5\right)}{a\left(b+5\right)+5\left(b+5\right)}+\dfrac{c\left(b+5\right)+5\left(c+5\right)}{b\left(c+5\right)+5\left(c+5\right)}+\dfrac{a\left(c+5\right)+5\left(a+5\right)}{a\left(c+5\right)+5\left(c+5\right)}\)
\(=\dfrac{b\left(a+5\right)+5\left(b+5\right)}{\left(a+5\right)\left(b+5\right)}+\dfrac{c\left(b+5\right)+5\left(c+5\right)}{\left(b+5\right)\left(c+5\right)}+\dfrac{a\left(c+5\right)+5\left(a+5\right)}{\left(a+5\right)\left(c+5\right)}\)
\(=\dfrac{b}{b+5}+\dfrac{5}{a+5}+\dfrac{c}{c+5}+\dfrac{5}{b+5}+\dfrac{a}{a+5}+\dfrac{5}{c+5}\)
\(=\left(\dfrac{b}{b+5}+\dfrac{5}{b+5}\right)+\left(\dfrac{a}{a+5}+\dfrac{5}{a+5}\right)+\left(\dfrac{c}{c+5}+\dfrac{5}{c+5}\right)\)
\(=1+1+1=3\) (\(a;b;c\ne-5\))
\(A=\dfrac{ab+5b+5b+25}{a\left(b+5\right)+5\left(b+5\right)}+\dfrac{bc+5c+5c+25}{b\left(c+5\right)+5\left(c+5\right)}+\dfrac{ca+5a+5a+25}{a\left(c+5\right)+5\left(c+5\right)}\)
\(A=\dfrac{b\left(a+5\right)+5\left(b+5\right)}{\left(a+5\right)\left(b+5\right)}+\dfrac{c\left(b+5\right)+5\left(c+5\right)}{\left(b+5\right)\left(c+5\right)}+\dfrac{a\left(c+5\right)+5\left(a+5\right)}{\left(a+5\right)\left(c+5\right)}\)
\(A=\dfrac{b}{b+5}+\dfrac{5}{a+5}+\dfrac{c}{c+5}+\dfrac{5}{b+5}+\dfrac{a}{a+5}+\dfrac{5}{c+5}\)
\(A=\dfrac{a+5}{a+5}+\dfrac{b+5}{b+5}+\dfrac{c+5}{c+5}=1+1+1=3\)
\(\dfrac{a+b}{3}=\dfrac{b+c}{4}=\dfrac{c+a}{5}\\ \Rightarrow\left\{{}\begin{matrix}4a+4b=3b+3c\\5a+5b=3c+3a\\5b+5c=4c+4a\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4a+b-3c=0\\4a-5b-c=0\\2a+5b-3c=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}4a+b-3c=0\\4a=5b+c\\3c=2a+5b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}5b+c+b-3c=0\\4a+b-2a-5b=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}6b=2c\\2a=4b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}c=3b\\a=2b\end{matrix}\right.\\ \Rightarrow M=20b+b-21b+2021=2021\)
=> 2(a+b+c)=10+25+25=60
=> a+b+c=60:2=30
các bạn cho mk vài li-ke cho tròn 700 với
tong do =30 nha
tick cho mình với mọi người