Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 2:
a: \(\Leftrightarrow\left(3x-5-2x-4\right)\left(3x-5+2x+4\right)=0\)
=>(x-9)(5x-1)=0
=>x=9 hoặc x=1/5
b: \(\Leftrightarrow\left(3x-2\right)\left(4x^2-1\right)=0\)
=>(3x-2)(2x-1)(2x+1)=0
hay \(x\in\left\{\dfrac{2}{3};\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
a) Ta có: \(A\left(x\right)=2x^5-3x^3+7x-6x^4+2x^3+2\)
\(=2x^5-6x^4-x^3+7x+2\)
Ta có: \(B\left(x\right)=x^5-3x^3+7x-6x^2+x^5+2x^2\)
\(=2x^5-3x^3-4x^2+7x\)
b) Ta có: \(A\left(x\right)-B\left(x\right)\)
\(=2x^5-6x^4-x^3+7x+2-\left(2x^5-3x^3-4x^2+7x\right)\)
\(=2x^5-6x^4-x^3+7x+2-2x^5+3x^3+4x^2-7x\)
\(=-6x^4+2x^3+4x^2+2\)
Ta có: \(A\left(x\right)+B\left(x\right)\)
\(=2x^5-6x^4-x^3+7x+2+2x^5-3x^3-4x^2+7x\)
\(=4x^5-6x^4-4x^3-4x^2+14x+2\)
c) Ta có: C(x)+2A(x)=B(x)
\(\Leftrightarrow C\left(x\right)=B\left(x\right)-2\cdot A\left(x\right)\)
\(\Leftrightarrow C\left(x\right)=2x^5-3x^3-4x^2+7x-2\cdot\left(2x^5-6x^4-x^3-7x+2\right)\)
\(\Leftrightarrow C\left(x\right)=2x^5-3x^3-4x^2+7x-4x^5+12x^4+2x^3+14x-4\)
\(\Leftrightarrow C\left(x\right)=-2x^5+12x^4-x^3-4x^2+21x-4\)
a: \(\Leftrightarrow\left(x+1\right)^2\left[\left(x+1\right)^2-1\right]=0\)
=>(x+1)^2(x+2)*x=0
hay \(x\in\left\{0;-1;-2\right\}\)
b: \(\Leftrightarrow\left(x-2\right)^3\left[\left(x-2\right)^2-1\right]=0\)
=>\(\left(x-2\right)^3\left(x-3\right)\left(x-1\right)=0\)
hay \(x\in\left\{2;3;1\right\}\)
c: \(\Leftrightarrow\left(2x-1\right)^2\left[\left(2x-1\right)^2-1\right]=0\)
=>\(\left(2x-1\right)^2\cdot2x\cdot\left(2x-2\right)=0\)
hay \(x\in\left\{0;\dfrac{1}{2};1\right\}\)
1)
a) \(2xy^2\left(x^2-2y\right)=2xy^2x^2-2xy^2\cdot2y=2x^3y^2-4xy^3\)
b) \(\left(x-3\right)\left(x+3\right)=x^2-3^2=x^2-9\)
c) \(\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-y^3\)
2)
a) \(2\left(x-4\right)-3\left(2x+7\right)=5\left(x-3\right)+12\) (1)
\(\Leftrightarrow2x-8-6x-21=5x-15+12\)
\(\Leftrightarrow2x-6x-5x=-15+12+8+21\)
\(\Leftrightarrow-9x=26\)
\(\Leftrightarrow x=-\dfrac{26}{9}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{-\dfrac{26}{9}\right\}\)
b) \(x\left(x+2\right)-x=2\) (2)
\(\Leftrightarrow x^2+2x-x=2\)
\(\Leftrightarrow x^2+x=2\)
\(\Leftrightarrow x^2+x-2=2-2\)
\(\Leftrightarrow x^2-x-2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy tập nghiệm phương trình (2) là \(S=\left\{-2;1\right\}\)
3)
\(2^m+2^n=2048\)
\(\Leftrightarrow2^m+2^n=2^8\)
\(\Leftrightarrow2^n\left(2^{m-n}-1\right)=2^8\)
Nếu:
♦ m - n = 0 (vô lý)
♦ m - n > 0:
\(\Rightarrow2^{m-n}-1\) lẻ mà \(2^8\) chẵn suy ra \(2^{m-n}-1=1\Rightarrow m=n+1\)
\(\Rightarrow2^n=2^8\Rightarrow n=8;m=9\)
Vậy \(n=8;m=9\)
Câu 1:
Ta có: \(M\left(x\right)=6x^3+2x^4-x^2+3x^2-2x^3-x^4+1-4x^3\)
\(=x^4+2x^2+1\)
\(=\left(x^2+1\right)^2\ge1\forall x\)
hay M(x) vô nghiệm(đpcm)
Câu 2:
Ta có: A(0)=5
\(\Leftrightarrow m+n\cdot0+p\cdot0\cdot\left(0-1\right)=5\)
\(\Leftrightarrow m=5\)
Ta có: A(1)=-2
\(\Leftrightarrow m+n\cdot1+p\cdot1\cdot\left(1-1\right)=-2\)
\(\Leftrightarrow5+n=-2\)
hay n=-2-5=-7
Ta có: A(2)=7
\(\Leftrightarrow5+\left(-7\right)\cdot2+p\cdot2\cdot\left(2-1\right)=7\)
\(\Leftrightarrow-9+2p=7\)
\(\Leftrightarrow2p=16\)
hay p=8
Vậy: Đa thức A(x) là 5-7x+8x(x-1)
\(=5-7x+8x^2-8x\)
\(=8x^2-15x+5\)
\(a)\left|x-1\right|=4 \Leftrightarrow\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
Vậy...
\(b)\left|x-1\right|+\left|x+3\right|=4\left(1\right)\)
*Khi \(x< -3\), phương trình (1) trở thành:
\(-\left(x-1\right)-\left(x+3\right)=4\\ \Leftrightarrow-x+1-x-3=4\\ \Leftrightarrow-2x-2=4\\ \Leftrightarrow-2x=6\\ \Leftrightarrow x=-3\left(KTM\right)\)
*Khi \(-3\le x< 1\), phương trình (1) trở thành:
\(-\left(x-1\right)+x+3=4\\ \Leftrightarrow-x+1+x+3=4\\ \Leftrightarrow0x+4=4\\ \Leftrightarrow0x=0\left(VSN\right)\)
*Khi \(x\ge1\), phương trình (1) trở thành:
\(x-1+x+3=4\\ \Leftrightarrow2x+2=4\\ \Leftrightarrow2x=2\\ \Leftrightarrow x=1\left(TM\right)\)
Vậy...
Câu 2:
a: THeo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}-a+b=-11\\2a+b=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=-9\end{matrix}\right.\)
b: Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}a\cdot0+b\cdot0+c=5\\4a-2b+c=21\\a-b+c=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=5\\4a-2b=16\\a-b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=5\\a=3\\b=-2\end{matrix}\right.\)
a) Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(\left|x-1\right|+\left|x+3\right|=\left|1-x\right|+\left|x+3\right|\ge\left|1-x+x+3\right|=4\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}1-x\ge0\\x+3\ge0\end{matrix}\right.\)
\(\Leftrightarrow-3\le x\le1\)
Vậy,..................................................................................................................................