a. -2x(x3 - 3x2 - x + 1)
= -2x4 + 6x3 + 2x2 - 2x
c. 3x2(2x3 - x + 5)
= 6x5 - 3x3 + 15x2
Bài 3:
a: Ta có: \(6x\left(5x-3\right)+3x\left(1-10x\right)=7\)
\(\Leftrightarrow30x^2-18x+3x-30x^2=7\)
\(\Leftrightarrow x=-\dfrac{7}{15}\)
b: Ta có: \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2+27x=30\)
hay x=2
c: ta có: \(x\left(5-2x\right)-2x\cdot\left(x-1\right)=15\)
\(\Leftrightarrow5x-2x^2-2x^2+2x-15=0\)
\(\Leftrightarrow-4x^2+7x-15=0\)
\(\text{Δ}=7^2-4\cdot\left(-4\right)\cdot\left(-15\right)=-191\)
Vì Δ<0 nên phương trình vô nghiệm
Bài 1:
f) \(-\dfrac{1}{3}xz\left(-9xy+15yz\right)+3x^2\left(2yz^2-yz\right)=3x^2yz-5xyz^2+6x^2yz^2-3x^2yz=6x^2yz^2-5xyz^2\)
Bài 3:
a) \(6x\left(5x+3\right)+3x\left(1-10x\right)=7\)
\(\Leftrightarrow30x+18x+3x-30x=7\Leftrightarrow21x=7\Leftrightarrow x=\dfrac{1}{3}\)
b) \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2+27x=30\Leftrightarrow15x=30\Leftrightarrow x=2\)
c) \(x\left(5+2x\right)-2x\left(x-1\right)=15\)
\(\Leftrightarrow5x+2x^2-2x^2+2x=15\Leftrightarrow7x=15\Leftrightarrow x=\dfrac{15}{7}\)
d) \(2x\left(3x-1\right)+\left(4-2x\right)3x=7\)
\(\Leftrightarrow6x^2-2x+12x-6x^2=7\)
\(\Leftrightarrow10x=7\Leftrightarrow x=\dfrac{7}{10}\)
f: ta có: \(-\dfrac{1}{3}xz\left(-9xy+15yz\right)+3x^2\left(2yz^2-yz\right)\)
\(=3x^2yz-5xyz^2+6x^2yz^2-3x^2yz\)
\(=-5xyz^2+6x^2yz^2\)
f. \(\dfrac{-1}{3}xz\left(-9xy+15yz\right)+3x^2\left(2yz^2-yz\right)\)
= \(3x^2yz-5xyz^2+6x^2yz^2-3x^2yz\)
= 3x2yz - 3x2yz - 5xyz2 + 6x2yz2
= 6x2yz2 - 5xyz2
Câu 1:
f. \(-\dfrac{1}{3}xz\left(-9xy+15yz\right)+3x^2\left(2yz^2-yz\right)\)
\(=3x^2yz-5xyz^2+6x^2yz^2-3x^2yz\)
\(=6x^2yz^2-5x^2yz\)
Câu 3: có bạn làm abc r nên mình làm d nha
d. \(2x\left(3x-1\right)+\left(4-2x\right)3x=7\)
\(\Leftrightarrow6x^2-2x+12x-6x^2=7\)
\(\Leftrightarrow10x=7\)
\(\Leftrightarrow x=\dfrac{7}{10}\)
