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a) \(...\Rightarrow x.\left(2+5\right)=14\Rightarrow x.7=14\Rightarrow x=14:7=2\)
b) \(...\Rightarrow x.\left(9+1\right)=20\Rightarrow x.10=20\Rightarrow x=20:10=2\)
c) \(...\Rightarrow x.\left(\dfrac{2}{3}+\dfrac{1}{3}\right)=1999\Rightarrow x.\dfrac{3}{3}=1999\Rightarrow x=1999\)
d) \(...\Rightarrow11.x+22=5.x+40\Rightarrow11.x-5.x=40-22\Rightarrow6.x=18\Rightarrow x=18:6=3\)
e) \(...\Rightarrow11.x-66=4.x+11\Rightarrow11.x-4.x=11+66\Rightarrow7.x=77\Rightarrow x=77:7=11\)
f) \(...\Rightarrow\left(3.x-12\right):x=12-10\)
\(\Rightarrow3.x-12=2.x\)
\(\Rightarrow3.x-2.x=12\)
\(\Rightarrow x=12\)
g) \(...\Rightarrow\left(5.x+7\right):x=26-20\)
\(\Rightarrow5.x+7=6.x\)
\(\Rightarrow6.x-5.x=7\)
\(\Rightarrow x=7\)
h) \(...\Rightarrow x.\left(1999-1\right)=1999.\left(1997+1\right)\)
\(\Rightarrow x.1998=1999.1998\)
\(\Rightarrow x=1999.1998:1998\)
\(\Rightarrow x=1999\)
a, \(x\times\) 2 + \(x\times\) 5 = 14
\(x\) \(\times\) ( 2 + 5) = 14
\(x\) \(\times\) 7 = 14
\(x\) = 14: 7
\(x\) = 2
b, \(x\times9\) + \(x\)= 20
\(x\) \(\times\)( 9 + 1) = 20
\(x\) \(\times\) 10 = 20
\(x\) = 2
c, \(x\) : \(\dfrac{3}{2}\) + \(x\times\dfrac{1}{3}\) = 1999
\(x\times\) \(\dfrac{2}{3}\) + \(x\) \(\times\dfrac{1}{3}\) = 1999
\(x\times\) ( \(\dfrac{2}{3}\) + \(\dfrac{1}{3}\)) = 1999
\(x\) = 1999
d, 11\(\times\)(\(x+2\)) = 5 \(\times\) \(x\) + 40
11 \(\times\) \(x\) + 22 = 5 \(\times\) \(x\) + 40
11 \(\times\) \(x\) = 5 \(\times\) \(x\) + 40 - 22
11 \(\times\) \(x\) = 5 \(\times\) \(x\) + 18
11 \(\times\) \(x\) - 5 \(\times\) \(x\) = 18
\(x\) \(\times\) ( 11 - 5) = 18
\(x\) \(\times\) 6 = 18
\(x\) = 3
a,Cách 1: (6/11 + 5/11 ) x 3/7 Cách 2 :(6/11 + 5/11) x 3/7
= 1 x 3/7 =6/11 x 3/7 + 5/11 x 3/7
= 3/7 = 18/77 + 15/77
= 3/7
b, Cách 1:3/5 x 7/9 - 3/5 x 2/9 Cách 2 :3/5 x 7/9 - 3/5 x 2/9
= 7/15 - 2/15 = 3/5 x (7/9 - 2/9 )
= 1/3 = 3/5 x 5/9
= 1/3
c, Cách 1:(6/7 - 4/7) : 2/5 Cách 2: ( 6/7- 4/7 ) : 2/5
= 2/7 : 2/5 = 6/7 : 2/5 - 4/7 : 2/5
= 5/7 = 15/7 - 10/7
= 5/7
d,Cách 1:8/15 : 2/11 + 7/15 : 2/11 Cách 2:8/15 : 2/11 +7/15 : 2/11
= 88/30 + 77/30 =(8/15+7/15) :2/11
= 11/2 = 1 : 2/11
= 11/2
Bài 2 cậu tự làm nhé !
a, \(\left(-7\right)\left(5-x\right)< 0\)
\(< =>5-x>0< =>x< 5\)
b, \(11⋮x-1< =>x-1\inƯ\left(11\right)\in\left\{-11;-1;1;11\right\}\) ( \(x\ne1\) )
\(x\in\left\{-10;0;2;12\right\}\)
c, \(x+8⋮x+1< =>x+1+7⋮x+1\)
\(< =>7⋮x+1< =>x+1\inƯ\left(7\right)\in\left\{-7;-1;1;7\right\}\left(x\ne-1\right)\)
\(< =>x\in\left\{-8;-2;0;6\right\}\)
d, \(\left(x+2\right)\left(5-x\right)>0\)
Chưa học lập bảng xét dấu nên xét TH em nhé !
Nhận thấy ( x + 2 ) ( 5 - x ) > 0 nên x + 2 và 5 - x phải cùng dấu
TH1 : \(\left\{{}\begin{matrix}x+2>0\\5-x>0\end{matrix}\right.< =>\left\{{}\begin{matrix}x>-2\\x< 5\end{matrix}\right.< =>-2< x< 5}\)
TH2:
\(\left\{{}\begin{matrix}x+2< 0\\5-x< 0\end{matrix}\right.< =>\left\{{}\begin{matrix}x< -2\\x>5\end{matrix}\right.< =>x\in\varnothing\)
Từ 2 TH ta kết luận { x | -2 < x < 5 }
Điều kiện về x là gì bạn? Số nguyên, số tự nhiên, số hữu tỉ,...?
a) Ta có: \(\left(x-2\right)\cdot x=2x\cdot\left(x+5\right)\)
\(\Leftrightarrow x\cdot\left(x-2\right)-2x\left(x+5\right)=0\)
\(\Leftrightarrow x\cdot\left[x-2-2\left(x+5\right)\right]=0\)
\(\Leftrightarrow x\left(x-2-2x-10\right)=0\)
\(\Leftrightarrow x\left(-x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)
Vậy: S={0;-8}
b) Ta có: \(\left(2x-5\right)\left(x+11\right)=\left(5-2x\right)\left(2x+1\right)\)
\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)-\left(5-2x\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)+\left(2x-5\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(x+11+2x+1\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(3x+12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3x+12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\3x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{5}{2};-4\right\}\)
c) Ta có: \(x^2+6x+9=4x^2\)
\(\Leftrightarrow\left(x+3\right)^2-\left(2x\right)^2=0\)
\(\Leftrightarrow\left(x+3-2x\right)\left(x+3+2x\right)=0\)
\(\Leftrightarrow\left(-x+3\right)\left(3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x+3=0\\3x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-3\\3x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy: S={3;-1}
d) Ta có: \(\left(x+2\right)\left(5-4x\right)=x^2+4x+4\)
\(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(5-4x-x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(-5x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\-5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\-5x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{3}{5}\right\}\)
a) \(A=x^2-3x-x+3+11\)
\(=\left(x^2-4x+4\right)+10\)
\(=\left(x-2\right)^2+10\ge10\forall x\in R\)
Dấu "=" xảy ra<=> \(\left(x-2\right)^2=0\Leftrightarrow x=2\)
b) \(B=5-4x^2+4x\)
\(=-\left(4x^2-4x+1\right)+6\)
\(=-\left(2x-1\right)^2+6\le6\forall x\in R\)
Dấu "=" xảy ra<=> \(-\left(2x-1\right)^2=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
c) \(C=\left(x^2-3x+1\right)\left(x^2-3x-1\right)\)
\(=\left(x^2-3x\right)^2-1\ge-1\forall x\in R\)
Dấu "=" xảy ra<=>\(\left(x^2-3x\right)^2=0\Leftrightarrow x\left(x-3\right)=0\Leftrightarrow x=0;x=3\)
Bạn viết rõ đề ra, \(x\inℕ\) hay \(x\inℤ\)?
Tiêu chí đề bài : để B,C,D là phân sô hay để B,C,D là số nguyên, số tự nhiên?