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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
Bài 1:
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{b}=\frac{c}{d}=\frac{3a}{3b}=\frac{2c}{2d}=\frac{3a+2c}{3b+2d}\)
\(\Rightarrow\frac{a}{b}=\frac{c}{d}=\frac{3a+2c}{3b+2d}\)
Vậy \(\frac{a}{b}=\frac{c}{d}=\frac{3a+2c}{3b+2d}\)
Bài 2:
Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk,c=dk\)
Ta có: \(\frac{ac}{bd}=\frac{bkdk}{bd}=k^2\) (1)
\(\frac{a^2-c^2}{b^2-d^2}=\frac{\left(bk\right)^2-\left(dk\right)^2}{b^2-d^2}=\frac{b^2.k^2-d^2.k^2}{b^2-d^2}=\frac{k^2.\left(b^2-d^2\right)}{b^2-d^2}=k^2\) (2)
Từ (1) và (2) suy ra \(\frac{ac}{bd}=\frac{a^2-c^2}{b^2-d^2}\)
Bài 3: Tương tự nhé bạn chỉ cần thay a = bk, c = dk vào thôi
a) ta có: \(M=\left(\frac{1}{3}a-\frac{1}{3}b\right)-\left(a+2b\right)\)
\(M=\frac{1}{3}a-\frac{1}{3}b-a-2b\)
\(M=(\frac{1}{3}a-a)+\left(\frac{-1}{3}b-2b\right)\)
\(M=\frac{-2}{3}a+\frac{-7}{3}b\)
\(N=\frac{1}{3}a-\frac{1}{3}b-\left(a-b\right)\)
\(N=\frac{1}{3}a-\frac{1}{3}b-a+b\)
\(N=\left(\frac{1}{3}a-a\right)+\left(b-\frac{1}{3}b\right)\)
\(N=\frac{-2}{3}a+\frac{2}{3}b\)
\(\Rightarrow M+N=\left(\frac{-2}{3}a+\frac{-7}{3}b\right)+\left(\frac{-2}{3}a+\frac{2}{3}b\right)\)
\(=\frac{-2}{3}a+\frac{-7}{3}b+\frac{-2}{3}a+\frac{2}{3}b\)
\(=\left(\frac{-2}{3}a-\frac{2}{3}a\right)+\left(\frac{-7}{3}b+\frac{2}{3}b\right)\)
\(=\frac{-4}{3}a+\frac{-5}{3}b\)
\(\Rightarrow M+N=\frac{-4}{3}a-\frac{5}{3}b\)
ta có: \(M-N=\left(\frac{-2}{3}a+\frac{-7}{3}b\right)-\left(\frac{-2}{3}a+\frac{2}{3}b\right)\)
\(=\frac{-2}{3}a+\frac{-7}{3}b+\frac{2}{3}a-\frac{2}{3}b\)
\(=\left(\frac{-2}{3}a+\frac{2}{3}a\right)+\left(\frac{-7}{3}b-\frac{2}{3}b\right)\)
\(=0+\frac{-10}{3}b=\frac{-10}{3}b\)
\(\Rightarrow M-N=\frac{-10}{3}b\)
b) ta có: \(M=2a^2+ab-b^2-\left(-a^2+b^2-ab\right)\)
\(M=2a^2+ab-b^2+a^2-b^2+ab\)
\(M=\left(2a^2+a^2\right)+\left(ab+ab\right)+\left(-b^2-b^2\right)\)
\(M=3a^2+2ab+\left(-2b^2\right)\)
\(N=3a^2+b^2-\left(ab-a^2\right)\)
\(N=3a^2+b^2-ab+a^2\)
\(N=\left(3a^2+a^2\right)+b^2-ab\)
\(N=4a^2+b^2-ab\)
rồi bn tính như mk phần a nha!
c) ta có: \(M=\left(x+cy-z\right)+y+x-\left(z-x-y\right)\)
\(M=x+cy-z+y+x-z+x+y\)
\(M=\left(x+x+x\right)+\left(y+y\right)+\left(-z-z\right)+cy\)
\(M=3x+2y+\left(-2z\right)+cy\)
\(N=x-\left(x-\left(y-z\right)-x\right)\)
\(N=x-\left(x-y+z-x\right)\)
\(N=x-x+y-z+x\)
\(N=\left(x-x+x\right)+y-z\)
\(N=x+y-z\)
bn tính giúp mk cộng trừ 2 đa thức M; N luôn nha! mk chỉ rút gọn cho bn thôi
CHÚC BN HỌC TỐT!!!!
Đặt\(\frac{a}{b}=\frac{c}{d}=k\left(k\in Q\right)\)
\(\Rightarrow\hept{\begin{cases}a=bk\left(1\right)\\c=dk\left(2\right)\end{cases}}\)
Ta lại có \(\frac{3a^2+c^2}{3b^2+d^2}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\left(3\right)\)
Thay \(\left(1\right),\left(2\right)vào\left(3\right)có\)
\(\frac{3b^2k^2+d^2k^2}{3b^2+d^2}=\frac{k^2\left(3b^2+d^2\right)}{3b^2+d^2}=k^2\left(4\right)\)
\(\frac{\left(a+c\right)^2}{\left(b+d\right)^2}=\frac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\frac{k^2\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\left(5\right)\)
Từ \(\left(4\right),\left(5\right)\Rightarrowđpcm\)
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