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Bài 1:
Đề sai bạn ơi, phải là A(x)=x3-2x2+x-5
a, \(A\left(x\right)+B\left(x\right)=x^3-2x^2+x-5-x^3+2x^2+3x-9\)\(=4x-16\)
\(A\left(x\right)-B\left(x\right)=x^3-2x^2+x-5+x^3-2x^2-3x+9\)\(=2x^3-4x^2-2x+4\)
b, \(A\left(x\right)+B\left(x\right)=4x-16=4\left(x-4\right)\)\(\Rightarrow x=4\)
Vậy nghiệm của A(x)+B(x) là 4
Bài 2:
a, \(C\left(x\right)=-8x^4+5x^4+2x^3-4x^3+x^2+x+5\)\(=-3x^4-2x^3+x^2+x+5\)
\(D\left(x\right)=3,5+x^4-4x^3-4x^3+7-2x^4-3x^5\)\(=-3x^5+x^4-2x^4-4x^3-4x^3+3.5+7\)
\(=-3x^5-x^4-8x^3+10,5\)
b, \(C\left(x\right)+D\left(x\right)=\)\(-3x^4-2x^3+x^2+x+5\)\(-3x^5-x^4-8x^3+10,5\)\(=-3x^5-4x^4-10x^3+x^2+x+15,5\)
\(Q\left(x\right)=\)\(C\left(x\right)-D\left(x\right)=\)\(-3x^4-2x^3+x^2+x+5\)\(+3x^5+x^4+8x^3-10,5\)
\(=3x^5-2x^4+6x^3+x^2+x-5,5\)
c, \(D\left(x\right)=\)\(-3x^5-x^4-8x^3+10,5\)(not ra)
a. \(x:\left(\dfrac{2}{3}\right)^4=\dfrac{2}{3}\)
\(\Rightarrow x:\dfrac{16}{81}=\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{2}{3}.\dfrac{16}{81}\)
\(\Rightarrow x=\dfrac{32}{243}\)
Vậy....
b. \(\left(\dfrac{-5}{3}\right)^2.x=\left(\dfrac{-5}{3}\right)^3\)
\(\Rightarrow x=\left(\dfrac{-5}{3}\right)^3:\left(\dfrac{-5}{3}\right)^2\)
\(\Rightarrow x=\dfrac{-5}{3}\)
Vậy....
c. \(x^2-0,25=0\)
\(\Rightarrow x^2=0+0,25\)
\(\Rightarrow x^2=0,25=\left(\pm0,5\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x=0,5\\x=-0,5\end{matrix}\right.\)
Vậy.........
a) \(a^3+a^2b-a^2c-abc=a^2\left(a+b\right)-ac\left(a+b\right)=a\left(a+b\right)\left(a-c\right)\)
b) mk chỉnh lại đề
\(x^2+2xy+y^2-xz-yz=\left(x+y\right)^2-z\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)\)
c) \(4-x^2-2xy-y^2=4-\left(x+y\right)^2=\left(2-x-y\right)\left(2+x+y\right)\)
d) \(x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
a: \(\left(2x+1\right)^2=\left(x-1\right)^2\)
=>2x+1=x-1 hoặc 2x+1=1-x
=>x=-2 hoặc x=0
b: \(\left(x^2-5\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow x\in\left\{\sqrt{5};-\sqrt{5};-3\right\}\)
c: \(3\left(x-1\right)\left(2x-1\right)=5\left(x+8\right)\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(6x-3-5x-40\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-43\right)=0\)
hay \(x\in\left\{1;43\right\}\)
d: \(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
=>x+1=0
hay x=-1
(5x + 1)2 = 36/49
=> (5x + 1)2 = (6/7)2
=> \(\orbr{\begin{cases}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{35}\\x=-\frac{13}{35}\end{cases}}\)
Làm từ phần b nha
b) \(\left(x-\frac{1}{9}\right)^3=\frac{2}{3}^6\)
\(\Rightarrow\left(x-\frac{2}{9}\right)^3=\left(\frac{1}{3}\right)^6\)
\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\frac{1^6}{3^6}\)
\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\frac{1}{3^6}\)
\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\frac{1}{729}\)
\(\Rightarrow x-\frac{2}{9}=\frac{1}{9}\)
\(x=\frac{1}{9}+\frac{2}{9}\)
\(x=\frac{3}{9}=\frac{1}{3}\)
c) Sai đề rồi, xem lại đi
d) \(\left(x-3,5\right)^2+\left(y-\frac{1}{10}\right)^4< 0\)
\(\Rightarrow\frac{10000y^4-4000y^3+600y^3-40y+10000x^2+122501-70000x}{10000}< 0\)
=> Sai \(\forall y\inℝ\)
(5x+2)(x-7)=0
suy ra 5x+2=0 hoặc x-7=0
5x = -2
x = -2/5 hoặc x=7
\(x^2-x-6=0\Rightarrow x^2-2x+3x-6\\ \Rightarrow x\left(x-2\right)+3\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+3\right)=0\)
hay x-2=0 hoặc x+3 = 0
vậy x = 2 hoặc x = -3
b) \(\left(x-3\right)^2=0,25\)
\(\Rightarrow\left(x-3\right)^2=\left(\pm0,5\right)^2\)
\(\Rightarrow x-3=\pm0,5.\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0,5\\x-3=-0,5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0,5+3\\x=\left(-0,5\right)+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3,5\\x=2,5\end{matrix}\right.\)
Vậy \(x\in\left\{3,5;2,5\right\}.\)
c) \(\left(2x-1,5\right)^3=\left(-0,27\right)\) (câu này sai đề rồi nhé).
d) \(3.2x+\left(-1,2\right).x+2,7=\left(-4,9\right)\)
\(\Rightarrow3.2x+\left(-1,2\right).x=\left(-4,9\right)-2,7\)
\(\Rightarrow3.2x+\left(-1,2\right).x=-7,6\)
\(\Rightarrow6.x+\left(-1,2\right).x=-7,6\)
\(\Rightarrow\left[6+\left(-1,2\right)\right].x=-7,6\)
\(\Rightarrow4,8.x=-7,6\)
\(\Rightarrow x=\left(-7,6\right):4,8\)
\(\Rightarrow x=-\frac{19}{12}\)
Vậy \(x=-\frac{19}{12}.\)
Chúc bạn học tốt!
b) (x-3)\(^2\) = 0,25
(x-3)\(^2\) = (0,5)\(^2\)
x-3 = 0,5
x = 0,5 + 3
x = 0,8