a) 8 ⋮ x và x >0
b) 12 ⋮ x và x > 0
c) -8 ⋮ x và 12 ⋮ x
d) x ⋮ 4 ; x ⋮ (-6) và -20 <x < -10
e) (x - 3). (y + 5) = -17
f) /2x + 1/ - 19 = -7
g) -28 – 7. /- 3x + 15/ = -70
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a) 4x – 20 = 0
⇔ 4x = 20
⇔ x = 20 : 4
⇔ x = 5
Vậy phương trình có nghiệm duy nhất x = 5.
b) 2x + x + 12 = 0
⇔ 3x + 12 = 0
⇔ 3x = -12
⇔ x = -12 : 3
⇔ x = -4
Vậy phương trình đã cho có nghiệm duy nhất x = -4
c) x – 5 = 3 – x
⇔ x + x = 5 + 3
⇔ 2x = 8
⇔ x = 8 : 2
⇔ x = 4
Vậy phương trình có nghiệm duy nhất x = 4
d) 7 – 3x = 9 – x
⇔ 7 – 9 = 3x – x
⇔ -2 = 2x
⇔ -2 : 2 = x
⇔ -1 = x
⇔ x = -1
Vậy phương trình có nghiệm duy nhất x = -1.
Em lớp 6 em chỉ làm dc phần a,b,c
Kết quả như sau:
a,4x-20=0
4x=20+0
4x=20
x=20:4
x=5
a. 2x+\(\dfrac{4}{5}\)=0 hoặc 3x-\(\dfrac{1}{2}\)=0
2x=- 4/5 hoặc 3x=1/2
x=-2/5 hoặc x=\(\dfrac{1}{6}\)
b. x-\(\dfrac{2}{5}\)=0 hoặc x+\(\dfrac{4}{7}\)=0
x=2/5 hoặc x=-\(\dfrac{4}{7}\)
d. x(1+5/8-12/16)=1
\(\dfrac{7}{8}\)x=1=> x=8/7
\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)
\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)
\(a,\Leftrightarrow\left(x+1-2x-4\right)\left(x+1+2x+4\right)=0\\ \Leftrightarrow\left(-x-3\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{5}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)^2+\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x+2+x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ c,ĐK:x\ge0\\ PT\Leftrightarrow x-3\sqrt{x}+4\sqrt{x}-12=0\\ \Leftrightarrow\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)=0\\ \Leftrightarrow\sqrt{x}=3\left(\sqrt{x}+4>0\right)\\ \Leftrightarrow x=9\left(tm\right)\)
Bài 3:
b: \(\Leftrightarrow x^2\left(x+1\right)^2=0\)
hay \(x\in\left\{0;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=0\)
=>x-1=0
hay x=1
d: \(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)
hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)
\(a,\left(-31\right).\left(x+7\right)=0\\ \Rightarrow x+7=0\\ \Rightarrow x=-7\\ b,\left(8-x\right).\left(x+13\right)=0\\ \Rightarrow\left[{}\begin{matrix}8-x=0\\x+13=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-13\end{matrix}\right.\\ c,\left(x^2-25\right)\left(3-x\right)=0\\ \Rightarrow\left(x-5\right)\left(x+5\right)\left(3-x\right)=0\\\Rightarrow \left[{}\begin{matrix}x-5=0\\x+5=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\\x=3\end{matrix}\right.\\ d,\left(x-3\right)\left(x^2+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x^2=-4\left(loại\right)\end{matrix}\right.\\ \Rightarrow x=3\)
`x/8 = 3/4 +(-5/8)`
`=>x/8 = 6/8 +(-5/8)`
`=>x/8 = 1/8`
`=>x=1`
`-----`
`x/12 =3/4 +(-2/3)`
`=>x/12 = 9/12 + (-8/12)`
`=> x/12=1/12`
`=>x=1`
`----`
`1+11/13=24/x`
`=> 13/13 +11/13=24/x`
`=> 24/13 =24/x`
`=>x=13`
`----`
`x/6 -3/4=1/12`
`=>x/6 = 1/12 +3/4`
`=>x/6 = 1/12 + 9/12`
`=>x/6 = 10/12`
`=>x/6= 5/6`
`=>x=5`
\(a,\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=0\\\dfrac{8}{5}+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{4}{5}\end{matrix}\right.\)
\(b,\dfrac{x-\dfrac{4}{7}}{x+\dfrac{1}{2}}>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{4}{7}>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{4}{7}< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{4}{7}\\x< -\dfrac{1}{2}\end{matrix}\right.\)
\(c,\dfrac{2x-3}{x+\dfrac{7}{4}}< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-3< 0\\x+\dfrac{7}{4}>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-3>0\\x+\dfrac{7}{4}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x >-\dfrac{7}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{7}{4}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-\dfrac{7}{4}< x< \dfrac{3}{2}\\x\in\varnothing\end{matrix}\right.\Leftrightarrow-\dfrac{7}{4}< x< \dfrac{3}{2}\)
a, Vì \(x>0\Rightarrow\)x là số nguyên dương, Vì \(8⋮x\Rightarrow x=Ư\left(8\right)=\left\{1;2:4;8\right\}\)
Vậy \(x=\left\{1;2;4;8\right\}\)
b, Vì \(x>0\Rightarrow\)x là số nguyên dương, Vì \(12⋮x\Rightarrow x=Ư\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
Vậy \(x=\left\{\dots\right\}\)
c, Vì \(x⋮-8,x⋮12\Rightarrow x=UC\left(-8,12\right)=\left\{\pm1;\pm2;\pm4\right\}\)
Vậy \(x=\left\{\dots\right\}\)
Các câu sau bạn làm tương tự nha. #hoctot