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Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0
a.
\(10⋮\left(x-1\right)\)
\(\Rightarrow x-1=Ư\left(10\right)\)
\(\Rightarrow x-1=\left\{-10;-5;-2;-1;1;2;5;10\right\}\)
\(\Rightarrow x=\left\{-9;-4;-1;0;2;3;6;11\right\}\)
b.
\(\left(x+5\right)⋮\left(x-2\right)\Rightarrow\left(x-2\right)+7⋮x-2\)
\(\Rightarrow7⋮x-2\)
\(\Rightarrow x-2=Ư\left(7\right)=\left\{-7;-1;1;7\right\}\)
\(\Rightarrow x=\left\{-5;1;3;9\right\}\)
c.
\(\left(3x+8\right)⋮\left(x-1\right)\)
\(\Rightarrow\left(3x-3+11\right)⋮\left(x-1\right)\)
\(\Rightarrow3\left(x-1\right)+11⋮x-1\)
\(\Rightarrow11⋮\left(x-1\right)\)
\(\Rightarrow x-1=Ư\left(11\right)=\left\{-11;-1;1;11\right\}\)
\(\Rightarrow x=\left\{-10;0;2;12\right\}\)
\(1\)) \(70:\frac{4x+720}{x}=\frac{1}{2}\)
\(\Leftrightarrow\frac{4x+720}{x}=70:\frac{1}{2}\)
\(\Leftrightarrow\frac{4x+720}{x}=140\)
\(\Leftrightarrow\left(4x+720\right):x=140\)
\(\Leftrightarrow4x+720=140.x\)
\(\Leftrightarrow4x-140x=-720\)
\(\Leftrightarrow x.\left(-136\right)=-720\)
\(\Leftrightarrow x=-720:\left(-136\right)\)
\(\Leftrightarrow x=\frac{90}{17}\)
\(2\)) Mình đang nghĩ
a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)
Th1 : \(x-\dfrac{1}{2}=0\)
\(x=0+\dfrac{1}{2}\)
\(x=\dfrac{1}{2}\)
Th2 : \(-3-\dfrac{x}{2}=0\)
\(\dfrac{x}{2}=-3\)
\(x=\left(-3\right)\cdot2\)
\(x=-6\)
Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)
b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)
\(x=\dfrac{5}{8}+\dfrac{1}{8}\)
\(x=\dfrac{3}{4}\)
c) \(-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2\)
\(\dfrac{3}{2}+x=-\dfrac{1}{2}-\left(-2\right)\)
\(\dfrac{3}{2}+x=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}-\dfrac{3}{2}\)
\(x=0\)
d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)
\(x+\dfrac{1}{3}=-4\)
\(x=-4-\dfrac{1}{3}\)
\(x=-\dfrac{13}{3}\)
a)\(10\left(x-7\right)-8\left(x+5\right)=6\cdot\left(-5\right)+24\)
\(10x-10\cdot7-8x-8\cdot5=\left(-30\right)+24\)
\(10x-70-8x-40=-6\)
\(10x-8x=\left(-6\right)+70+40\)
\(2x=104\)
\(x=104\div2\)
\(x=52\)
b)\(2\left(4x-8\right)-7\left(3+x\right)=6\)
\(2\cdot4x-2\cdot8-7\cdot3-7x=6\)
\(8x-16-21-7x=6\)
\(8x-7x=6+16+21\)
\(x=43\)
Bài 8:
a) Đk: x#2 (*)
Với (*), A=(x - 2 + 5)/(x - 2)= 1 + 5/(x - 2)
A nguyên <=> x-2 thuộc Ư(5)={-5;-1;1;5}
=> S={-3;1;3;7}
b) Đk: x#-3
Với (*), A= (- 2x - 6 + 7)/(x + 3) = -2 + 7/(x+3)
A nguyên <=> x + 3 thuộc Ư(7)={1;-1;7;-7}
=> S = {-2;- 4;4;-10}
) Đk: x#2 (*)
Với (*), A=(x - 2 + 5)/(x - 2)= 1 + 5/(x - 2)
A nguyên <=> x-2 thuộc Ư(5)={-5;-1;1;5}
=> S={-3;1;3;7}
b) Đk: x#-3
Với (*), A= (- 2x - 6 + 7)/(x + 3) = -2 + 7/(x+3)
A nguyên <=> x + 3 thuộc Ư(7)={1;-1;7;-7
10 - 2(x+3) = 5(x - 2)
\(\Rightarrow\)10 - (2x+6) = 5x -10
\(\Rightarrow\)10 - 2x - 6 = 5x -10
\(\Rightarrow\)10 - 6 + 10 = 5x + 2x
\(\Rightarrow\)14=7x
\(\Rightarrow\)x = 14:7=2
Vậy x=2
\(\Leftrightarrow10-2x-6=5x-10\)
\(\Leftrightarrow-2x-5x=-10-10+6\)
\(\Leftrightarrow-7x=-14\)
\(\Leftrightarrow x=2\)