tìm x để : (x+6):(x+4)
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Hôm nay olm.vn sẽ hướng dẫn các em cách giải dạng bài như này.
Gặp những dạng toán nâng cao như này thì các em cần tìm \(x\) dưới dạng tổng quát em nhé. Học toán tập hợp là để giải toán dạng này đó em
Bài 3: a, 12 + 36 + 24 + \(x\) = 72 + \(x\)
72 + \(x\) ⋮ 6 ⇔ \(x\) ⋮ 6 ⇒ \(x\in\) A = { \(x\in\) Z/ \(x\) = 6k; k \(\in\) Z}
b, 72 + \(x\) không chia hết cho 6 ⇒ \(x\) không chia hết cho 6
⇒ \(x\) \(\in\) A = { \(x\) \(\in\) z/ \(x\) = 6k + q; k \(\in\) Z; q \(\in\) Z; q \(\ne\)0}
Bài 4: \(x\).9 ⋮3 vì 9 ⋮ 3 ⇒ \(x.9\) ⋮ 3 ∀ \(x\) \(\in\) Z Vậy \(x\) \(\in\) Z
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\(a,P=\left(\dfrac{\sqrt{x}-1}{x-4}-\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\right).\dfrac{x\sqrt{x}-2x-4\sqrt{x}+8}{6\sqrt{x}-18}\left(dk:x\ne4,x\ge0,x\ne9\right)\)
\(=\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)^2}\right).\dfrac{\sqrt{x^2}\left(\sqrt{x}-2\right)-4\left(\sqrt{x}-2\right)}{6\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)^2\left(\sqrt{x}+2\right)}.\dfrac{\left(x-4\right)\left(\sqrt{x}-2\right)}{6\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-3\sqrt{x}+2-x-3\sqrt{x}-2}{\left(x-4\right)\left(\sqrt{x}-2\right)}.\dfrac{\left(x-4\right)\left(\sqrt{x}-2\right)}{6\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-6\sqrt{x}}{6\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-\sqrt{x}}{\sqrt{x}-3}\)
\(b,P>0\Leftrightarrow\dfrac{-\sqrt{x}}{\sqrt{x}-3}>0\Leftrightarrow-\sqrt{x}>0\Leftrightarrow\sqrt{x}< -1\left(ktm\right)\)
\(\Leftrightarrow\sqrt{x}-3>0\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)
\(c,P< 1\Leftrightarrow-\dfrac{\sqrt{x}}{\sqrt{x}-3}< 1\Leftrightarrow-\sqrt{x}< 1\Leftrightarrow\sqrt{x}>-1\left(ktm\right)\)
\(\Leftrightarrow\sqrt{x}-3< 1\Leftrightarrow\sqrt{x}< 4\Leftrightarrow x< 2\)
a: \(P=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)^2\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)-2\sqrt{x}\left(\sqrt{x}+2\right)}{6\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)^2}\cdot\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)^2}{6\left(\sqrt{x}-3\right)}\)
=1/3(căn x-3)
b: P>0
=>căn x-3>0
=>x>9
c: P<1
=>P-1<0
=>\(\dfrac{1-3\sqrt{x}+9}{3\sqrt{x}-9}< 0\)
=>\(\dfrac{-3\sqrt{x}+10}{3\sqrt{x}-9}< 0\)
=>(3căn x-10)/(3căn x-9)>0
=>x>100/3 hoặc 0<x<9
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a: \(A=\dfrac{x^2-5x+6-x^2+x+2x^2-6}{x\left(x-3\right)}=\dfrac{2x^2-4x}{x\left(x-3\right)}=\dfrac{2x}{x-3}\)
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a: \(Q=-\dfrac{7}{12}xy^2+\dfrac{4}{3}x-\dfrac{1}{2}x^2y-1\)
\(A=x^2y-3x+1-\dfrac{7}{12}xy^2+\dfrac{4}{3}x-\dfrac{1}{2}x^2y-1=\dfrac{1}{2}x^2y-\dfrac{7}{12}xy^2-3x\)
b: \(P=\dfrac{3}{4}xy^2+\dfrac{4}{9}x-\dfrac{7}{12}xy^2+\dfrac{4}{3}x-\dfrac{1}{2}x^2y-1=\dfrac{1}{6}xy^2+\dfrac{16}{9}x-\dfrac{1}{2}x^2y-1\)
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(2 + x) + (4 + x) + .... + (52 + x) = 780
(2 + 4 + 6 + .... + 52) + (x + x + x + .... + x) = 780
26.(52 + 2) :2 + 26x = 780
702 + 26x = 780
26x = 780 - 702
26x = 78
x = 78 : 26
x = 3
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1: Ta có: \(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-9-\left(x-9\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
Để \(A=-\dfrac{1}{\sqrt{x}}\) thì \(x+\sqrt{x}=-\sqrt{x}+3\)
\(\Leftrightarrow x+2\sqrt{x}-3=0\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow x=1\left(nhận\right)\)
2: Để A nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{-1;1;2;-2;4;-4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{2;4;5;1;7\right\}\)
\(\Leftrightarrow x\in\left\{16;25;1;49\right\}\)
\(x+6⋮x+4\)
\(\Rightarrow x+4+2⋮x+4\)
\(\Rightarrow2⋮x+4\)
\(\Rightarrow x+4\in\left(1;-1;2;-2\right)\)
\(\Rightarrow x\in\left(-3;-5;-2;-6\right)\)
( x + 6 ) chia hết cho ( x + 4 )
=> ( x + 4 ) + 2 chia hết cho ( x + 4 )
do : ( x + 4 ) chia hết ( x + 4 )
để : ( x + 6 ) chia hết cho ( x + 4 ) => 2 phải chia hết cho x + 4
=> x + 4 thuộc Ư(2) = { 1 ; 2 }
nếu x + 4 = 1 thì x = -3
nếu x + 4 = 2 thì x = -2
x = { -3 ; - 2 }