Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a) -6x + 3(7 + 2x)
= -6x + 21 + 6x
= (-6x + 6x) + 21
= 21
b) 15y - 5(6x + 3y)
= 15y - 30 - 15y
= (15y - 15y) - 30
= -30
c) x(2x + 1) - x2(x + 2) + (x3 - x + 3)
= 2x2 + x - x3 - 2x2 + x3 - x + 3
= (2x2 - 2x2) + (x - x) + (-x3 + x3) + 3
= 3
d) x(5x - 4)3x2(x - 1) ??? :V
Bài 2:
a) 3x + 2(5 - x) = 0
<=> 3x + 10 - 2x = 0
<=> x + 10 = 0
<=> x = -10
=> x = -10
b) 3x2 - 3x(-2 + x) = 36
<=> 3x2 + 2x - 3x2 = 36
<=> 6x = 36
<=> x = 6
=> x = 5
c) 5x(12x + 7) - 3x(20x - 5) = -100
<=> 60x2 + 35x - 60x2 + 15x = -100
<=> 50x = -100
<=> x = -2
=> x = -2
\(\left(x-1\right)^3=27\)
\(\Leftrightarrow\left(x-1\right)^3=3^3\)
\(\Leftrightarrow x-1=3\)
\(\Leftrightarrow x=4\)
\(x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
Vậy x = 0 hoặc x = -1
\(\left(2x+1\right)^2=25\)
\(\Leftrightarrow\left(2x+1\right)^2=\left(\pm5\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
Vậy x = 2 hoặc x = -3
\(\left(2x-3\right)^2=36\)
\(\Leftrightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=6\\2x-3=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4,5\\x=-1,5\end{cases}}\)
Vậy x = 4,5 hoặc x = -1,5
Bài 1:
a)
\(\dfrac{4^2\cdot25^2+32\cdot125}{2^3\cdot5^2}\\ =\dfrac{\left(2^2\right)^2\cdot\left(5^2\right)^2+2^5\cdot5^3}{2^3\cdot5^2}\\ =\dfrac{2^{2\cdot2}\cdot5^{2\cdot2}+2^5\cdot5^3}{2^3\cdot5^2}\\ =\dfrac{2^4\cdot5^4+2^5\cdot5^3}{2^3\cdot5^2}\\ =\dfrac{2^4\cdot5^4}{2^3\cdot5^2}+\dfrac{2^5\cdot5^3}{2^3\cdot5^2}\\ =2\cdot5^2+2^2\cdot5\\ =2\cdot25+4\cdot5\\ =50+20\\ =70\)
c)
\(\dfrac{\left(1-\dfrac{4}{9}-2\right)\cdot16}{\left(2-3\right)^{-2}}+12\\ =\dfrac{\left(\dfrac{9}{9}-\dfrac{4}{9}-\dfrac{18}{9}\right)\cdot16}{\left(-1\right)^{-2}}+12\\ =\dfrac{\dfrac{-13}{9}\cdot16}{\dfrac{1}{\left(-1\right)^2}}+12\\ =\dfrac{\dfrac{-208}{9}}{1}+12\\ =\dfrac{-208}{9}+12\\ =\dfrac{-208}{9}+\dfrac{108}{9}\\ =\dfrac{100}{9}\)
Bài 2:
a)
\(\left(x+2\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)
b)
\(\left(1,78^{2x-2}-1,78^x\right):1,78^x=0\\ \Leftrightarrow\dfrac{1,78^{2x-2}}{1,78^x}-\dfrac{1,78^x}{1,78^x}=0\\ \Leftrightarrow\dfrac{1,78^{2x-2}}{1,78^x}-1=0\\ \Leftrightarrow \dfrac{1,78^{2x-2}}{1,78^x}=1\\ \Leftrightarrow1,78^{2x-2}=1,78^x\\ \Leftrightarrow2x-2=x\\ \Leftrightarrow2x-x=2\\ \Leftrightarrow x=2\)
d) \(5^{\left(x-2\right)\left(x+3\right)}=1\)
\(\Rightarrow5^{\left(x-2\right)\left(x+3\right)}=5^0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy \(x_1=-3;x_2=2\)
Căng, sự thật là nó rất căng
Nhg dù sao thì.....
1) \(A\left(x\right)=\left(x-4\right)^2-\left(2x+1\right)^2\)
Xét \(A\left(x\right)=0\)
\(\Rightarrow\left(x-4\right)^2-\left(2x+1\right)^2=0\)
\(\Rightarrow x^2-8x+16-4x^2-4x-1=0\)
\(\Rightarrow-3x^2-12x+15=0\)
\(\Rightarrow-3x^2+3x-15x+15=0\)
\(\Rightarrow-3x\left(x-1\right)-15\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(-3x-15\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\-3x-15=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
2)(Sửa đề nha, sai cmnr) \(B\left(x\right)=x^3+x^2-4x-4\)
Xét \(B\left(x\right)=0\)
\(\Rightarrow x^3+x^2-4x-4=0\)
\(\Rightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\)
\(\Rightarrow\left(x^2-4\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\pm2\\x=-1\end{matrix}\right.\)
Đó là những j mình biết 
1/a, f(x) - g(x) + h(x) = x3 - 2x2 + 3x +1 - x3 - x + 1 +2x2 - 1
=(x3 - x3) + (-2x2 + 2x2) + (3x - x) + (1 + 1 - 1)
=2x + 1
b, f(x) - g(x) + h(x) = 0
<=> 2x + 1 = 0
<=> 2x = -1
<=> x = -1/2
Vậy x = -1/2 là nghiệm của đa thức f(x) - g(x) + h(x)
2/ a, 5x + 3(3x + 7)-35 = 0
<=> 5x + 9x + 21 - 35 = 0
<=> 14x - 14 = 0
<=> 14(x - 1) = 0
<=> x-1 = 0
<=> x = 1
Vậy 1 là nghiệm của đa thức 5x + 3(3x + 7) -35
b, x2 + 8x - (x2 + 7x +8) -9 =0
<=> x2 + 8x - x2 - 7x - 8 - 9 =0
<=> (x2 - x2) + (8x - 7x) + (-8 -9)
<=> x - 17 = 0
<=> x =17
Vậy 17 là nghiệm của đa thức x2 + 8x -(x2 + 7x +8) -9
3/ f(x) = g (x) <=> x3 +4x2 - 3x + 2 = x2(x + 4) + x -5
<=> x3 +4x2 - 3x + 2 = x3 + 4x2 + x - 5
<=> -3x + 2 = x - 5
<=> -3x = x - 5 - 2
<=> -3x = x - 7
<=>2x = 7
<=> x = 7/2
Vậy f(x) = g(x) <=> x = 7/2
4/ có k(-2) = m(-2)2 - 2(-2) +4 = 0
=> 4m + 4 + 4 = 0
=> 4m + 8 = 0
=> 4m = -8
=> m = -2