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Bài 1:
a)
\(F+G+H=(x^3-2x^2+3x+1)+(x^3+x-1)+(2x^2-1)\)
\(=2x^3+4x-1\)
b)
\(F-G+H=0\)
\(\Leftrightarrow (x^3-2x^2+3x+1)-(x^3+x-1)+(2x^2-1)=0\)
\(\Leftrightarrow 2x+1=0\)
\(\Leftrightarrow x=-\frac{1}{2}\)
Bài 2:
a)
\(A=-4x^5-x^3+4x^2-5x+9+4x^5-6x^2-2\)
\(=(-4x^5+4x^5)-x^3+(4x^2-6x^2)-5x+(9-2)\)
\(=-x^3-2x^2-5x+7\)
\(B=-3x^4-2x^3+10x^2-8x+5x^3\)
\(=-3x^4+(5x^3-2x^3)+10x^2-8x\)
\(=-3x^4+3x^3+10x^2-8x\)
b)
\(P=A+B=(-x^3-2x^2-5x+7)+(-3x^4+3x^3+10x^2-8x)\)
\(=-3x^4+(3x^3-x^3)+(10x^2-2x^2)-(8x+5x)+7\)
\(=-3x^4+2x^3+8x^2-13x+7\)
\(P(-1)=-3.(-1)^4+2(-1)^3+8(-1)^2-12(-1)+7=23\)
\(Q=A-B=(-x^3-2x^2-5x+7)-(-3x^4+3x^3+10x^2-8x)\)
\(=3x^4-(x^3+3x^3)-(2x^2+10x^2)+(8x-5x)+7\)
\(=3x^4-4x^3-12x^2+3x+7\)
a. Ta có: P=\(-3x^2+x-5x^3+2x+x^2+3x^3-4\)4
\(P=-2x^3-2x^2+3x-5\)
\(Q=7x^3-3-x-3x^2-4x-4x^3+x^2-3x^3\)
\(\Rightarrow Q=-2x^2-5x-3\)
a) (2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)
<=> 2x2 - 5x - 12 + x2 - 7x + 10 = 3x2 - 17x + 20
<=> 5x = 22
<=> x = 22/5
b) (8x - 3)(3x+ 2) - (4x + 7)(x + 4) = (2x + 1)(5x- 1)
<=> 24x2 + 7x - 6 - 4x2 - 23x - 28 = 10x2 + 3x - 1
<=> 10x2 - 19x -33 = 0
<=> 10x2 - 30x + 11x - 33 = 0
<=> (10x + 11)(x - 3) = 0
<=> \(\orbr{\begin{cases}10x+11=0\\x-3=0\end{cases}}\) <=> \(\orbr{\begin{cases}x=-\frac{11}{10}\\x=3\end{cases}}\)
c) 2x2 + 3(x - 1)(x + 1) = 5x(x + 1)
<=> 2x2 + 3x2 - 3 = 5x2 + 5x
<=> 5x = -3
<=> x = -3/5
a: \(\Leftrightarrow2^x\cdot2\cdot2^{-2}\cdot5=384\)
\(\Leftrightarrow2^x\cdot\dfrac{5}{2}=384\)
\(\Leftrightarrow2^x=153,6\)(vô lý)
b: Sửa đề:\(3^{x+2}\cdot5^y=45^x\)
\(\Leftrightarrow3^{x+2}\cdot5^y=3^{2x}\cdot5^x\)
=>x=y và 2x=x+2
=>x=y=2
B(x) - Q(x) = P(x)
=>B(x) = P(x) + Q(x)
=>B(x) = (2x2-5x+x3-1) + (x3+2x2-7+5x)
=>B(x) = 2x2-5x+x3-1 + x3+2x2-7+5x
=>B(x) = 2x2+2x2-5x+5x+x3+x3-1-7
=>B(x) = (2x2+2x2)-(5x-5x)+(x3+x3)-(1+7)
=>B(x) = 4x2-0+2x3-8
=>B(x) = 4x2+2x3-8
Bài 1 Từ \(\left|x-1\right|=2\) \(\Rightarrow\left\{{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=2+1\\x=-2+1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Với x = 3 \(\Rightarrow A=3\cdot3^2-4\cdot3+1\)
\(A=3\cdot9-12+1\)
\(A=27-12+1=16\)
Với x = -1 \(\Rightarrow A=3\cdot\left(-1\right)^2-4\cdot\left(-1\right)+1\)
\(A=3\cdot1-\left(-4\right)+1\)
\(A=3+4+1=8\)
Bài 2
a) \(\left|x-10\right|=-x+10=10-x\)
\(\left|x-10\right|=10-x\) \(\Leftrightarrow x-10\le0\) \(\Leftrightarrow x\le10\)
b)\(\left|\dfrac{x-1}{1-x}\right|=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{1-x}=1\\\dfrac{x-1}{1-x}=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-1=1-x\\x-1=-1+x\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x+x=1+1\\x-x=-1+1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}2x=2\\0=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2:2=1\\x\in Z\end{matrix}\right.\)
d)\(\left|x-4\right|+\left(3+3\right)=7\)
\(\left|x-4\right|+6=7\)
\(\left|x-4\right|=7-6\)
\(\left|x-4\right|=1\)
\(\Rightarrow\left\{{}\begin{matrix}x-4=1\\x-4=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1+4\\x=-1+4\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)