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bài 1 .
a. 3 x(5x2 – 2x -1) = 15x3 – 6x2 – 3x
b. (x2+2xy -3)(-xy) = – x3y – 2x2y2 + 3xy
c. 1/2 x2y ( 2x3 – 2/5 xy2 -1 )= x5y – 1/5 x3y3 – 1/2 x2y
bài 2 .
a) 2x^3-3x-5x^3-x^2+x^2=-3x-3x^3
b) 3x^2-6x-5x+5x^2-8x^2+24=-11x+24
c) 3x^3-3/2x^2-x^3-x/2+x/2+2=2x^3-3/2x^2+2
bài 3 .
?????????? bài 3 thì tui ko biết
Bài 3 :
\(P=5x\left(x^2-3\right)+x^2\left(7-5x\right)-7x^2\)
\(=5x^3-15x+7x^2-5x^3-7x^2=-15x\)
Thay x = -5 vào biểu thức trên ta được
\(-15.\left(-5\right)=75\)
Vậy x = -5 thì P = 75
\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x=-\frac{13}{4}\)
\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)
\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)
\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)
\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)
\(\Leftrightarrow x=-\frac{6}{11}\)
d,e,f Tương tự
a: \(\Leftrightarrow12x^2-10x-12x^2-28x=7\)
=>-38x=7
hay x=-7/38
b: \(\Leftrightarrow-10x^2-5x+9x^2+6x+x^2-\dfrac{1}{2}x=0\)
=>1/2x=0
hay x=0
c: \(\Leftrightarrow18x^2-15x-18x^2-14x=15\)
=>-29x=15
hay x=-15/29
d: \(\Leftrightarrow x^2+2x-x-3=5\)
\(\Leftrightarrow x^2+x-8=0\)
\(\text{Δ}=1^2-4\cdot1\cdot\left(-8\right)=33>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{33}}{2}\\x_2=\dfrac{-1+\sqrt{33}}{2}\end{matrix}\right.\)
e: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)
\(\Leftrightarrow-25x^2=4\)
\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)
a) | \(\frac{1}{2}\)x| = 3 - 2x
\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}x=3-2x\\\frac{1}{2}x=-\left(3-2x\right)\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{1}{2}x+2x=3\\\frac{1}{2}x=-3+2x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{2}x=3\\\frac{1}{2}x-2x=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=3:\frac{5}{2}\\-\frac{3}{2}x=-3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{6}{5}\\x=-3:\left(-\frac{3}{2}\right)\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{6}{5}\\x=2\end{cases}}\)
b) |x - 1| = 3x + 2
\(\Rightarrow\orbr{\begin{cases}x-1=3x+2\\x-1=-\left(3x+2\right)\end{cases}}\Rightarrow\orbr{\begin{cases}x-3x=2+1\\x-1=-3x-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-2x=3\\x+3x=-2+1\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{-2}\\4x=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=-\frac{1}{4}\end{cases}}\)
c) | 5x | = x - 12
\(\Rightarrow\orbr{\begin{cases}5x=x-12\\5x=-\left(x-12\right)\end{cases}}\Rightarrow\orbr{\begin{cases}5x-x=-12\\5x=-x+12\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}4x=-12\\5x+x=12\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\6x=12\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
d) |7 - x| = 5x + 1
\(\Rightarrow\orbr{\begin{cases}7-x=5x+1\\7-x=-\left(5x+1\right)\end{cases}}\Rightarrow\orbr{\begin{cases}7-1=5x+x\\7-x=-5x-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}6=6x\\7+1=-5x+x\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\8=-4x\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)
e) |9 + x| = 2x
\(\Rightarrow\orbr{\begin{cases}9+x=2x\\9+x=-2x\end{cases}}\Rightarrow\orbr{\begin{cases}9=2x-x\\9=-2x-x\end{cases}}\Rightarrow\orbr{\begin{cases}9=x\\9=-3x\end{cases}}\Rightarrow\orbr{\begin{cases}x=9\\x=-3\end{cases}}\)
Ủng hộ mk nha !!! ^_^
1, \(\left(x-4\right)^2-\left(2x+1\right)^2=\left(x-4-2x-1\right)\left(x-4+2x+1\right)=-3\left(x+5\right)\left(x-1\right).\)
\(\orbr{\begin{cases}x+5=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=1\end{cases}}}\)(mấy cái này áp dụng hàng đẳng thức lớp 8 mới hok)
2,\(x^3+x^2-4x-4=\left(x-2\right)\left(x^2+3x+2\right)=\left(x-2\right)\left(x+1\right)\left(x+2\right)\)
\(\orbr{\begin{cases}x=\mp2\\\end{cases}}x=-1\)
tương tụ lm tiếp nhe buồn ngủ quá rồi !
ta có: f(x) + g(x) = ( 7 x^6 - 6x ^5 +5x^4 -4x^3 +3x^2 -2x +1) - ( x - 2x^2 +3x^3 - 4x^4 + 5x^5 - 6x^6)
\(=7x^6-6x^5+5x^4-4x^3+3x^2-2x+1-x+2x^2-3x^3+4x^4-5x^5+6x^6\)
\(=\left(7x^6+6x^6\right)-\left(6x^5+5x^5\right)+\left(5x^4+4x^4\right)-\left(4x^3+3x^3\right)+\left(3x^2+2x^2\right)-\left(2x+x\right)+1\)
\(=13x^6-11x^5+9x^4-7x^3+5x^2-3x+1\)
Chúc bn học tốt !!!!!!
Uhhhhhhhhhhhhhhhhhhhhhhhhhh😥😥😥😥😥😥😥😥😥😥😥????????????...............
