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Bài 2 :
a ) a - ( b + a ) = a - b - a = a - a - b = -b
b ) ( a + b + c ) - ( a + b - c ) = a + b + c - a - b - c = ( a - a ) + ( b - b ) + ( c - c ) = 0
c ) ( a + b - c ) + ( a - b + c ) - ( b + c - a ) = a + b - c + a - b + c - b - c + a = ( a + a + a ) + ( b - b - b ) + ( -c + c -c ) = a3 - b - c
Bài 3.
a, \(\left(-12+x\right)\left(x-9\right)< 0\)
TH1:\(\left\{{}\begin{matrix}-12+x>0\\x-9< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>12\\x< 9\end{matrix}\right.\)(vô lý)
TH2:\(\left\{{}\begin{matrix}-12+x< 0\\x-9>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< 12\\x>9\end{matrix}\right.\)\(\Rightarrow9< x< 12\)
Vậy \(9< x< 12\) thì thỏa mãn đề
b, \(\left(11-x^2\right)\left(45-x^2\right)>0\)
TH1:\(\left\{{}\begin{matrix}11-x^2>0\\45-x^2>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x^2< 11\\x^2< 45\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< \sqrt{11}\\x< \sqrt{45}\end{matrix}\right.\) \(\Rightarrow x< \sqrt{11}\)
TH2:\(\left\{{}\begin{matrix}11-x^2< 0\\45-x^2< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x^2>11\\x^2>45\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>\sqrt{11}\\x>\sqrt{45}\end{matrix}\right.\) \(\Rightarrow x>\sqrt{45}\)
Vậy \(x< \sqrt{11}\) hoặc \(x>\sqrt{45}\)
Bài 5,
a/ \(\left(2x+2\right)\left(2y-1\right)=23\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+2\inƯ\left(23\right)\\2y-1\inƯ\left(23\right)\end{matrix}\right.\)
Ta có bảng:
| 2x+2 | -23 | -1 | 1 | 23 |
| 2y-1 | -1 | -23 | 23 | 1 |
| x | \(\dfrac{-25}{2}\)(loại) | \(\dfrac{-3}{2}\)(loại) | \(\dfrac{-1}{2}\)(loại) | \(\dfrac{21}{2}\) (loại) |
| y | 0 | -11 | 12 | 1 |
Vậy k có cặp (x;y) nào tm yêu cầu của đề bài
b,c tương tự
Bai 1:a)54; b)-65.
Bai 2: a)-1004; b)13.
Bai 3:a)-38; B)-139; C) 313.
Bai 4:chiu.
k cho minh nha
1a) (2003 - 2003) + (75 - 21) = 54
b) 1152 - 374 - 1152 - 65 + 374 = (1152 - 1152) + (374 - 374) - 65 = -65
mk chỉ biết câu 1 thôi xin lỗi vì mk ko thể nào giải hết được !!!!!!!!!!!!!!!!!!!!!!!!
SORRY
a, A =(a-2b+c)-(a-2b-c)
A = a-2b+c-a+2b+c
A = 0
b, B = (-x-y+3)-(-x+2-y)
B = -x-y+3+x-2+y
B = 1
c, C = 2.(3a+b-1)-3.(2a+b-2)
C = 6a+2b-2-6a-3b+6
C = -b + 4
d, D = 4.(x-1)-(3x+2)
D = 4x-4-3x-2
D = x-6
Bài 1:
a) \(\frac{16}{15}.\frac{\left(-5\right)}{14}.\frac{54}{24}.\frac{56}{21}\)
\(=\frac{4.2.2}{5.3}.\frac{\left(-5\right)}{2.7}.\frac{3.3}{4}.\frac{8}{3}\)
\(=\frac{4.2.2.\left(-5\right).3.3.8}{5.3.2.7.4.3}\)
\(=\frac{-16}{7}\)
b) \(\frac{7}{3}.\frac{\left(-5\right)}{2}.\frac{15}{21}.\frac{4}{\left(-5\right)}\)
\(=\frac{7}{3}.\frac{\left(-5\right)}{2}.\frac{5}{7}.\frac{2.2}{\left(-5\right)}\)
\(=\frac{7.\left(-5\right).5.2.2}{3.2.7.\left(-5\right)}\)
\(=\frac{10}{3}\)
Bài 2:
a) \(\frac{21}{24}.\frac{11}{9}.\frac{5}{7}=\frac{7}{8}.\frac{11}{9}.\frac{5}{7}=\frac{11.5}{8.9}=\frac{55}{72}\)
b) \(\frac{5}{23}.\frac{17}{26}+\frac{5}{23}.\frac{9}{26}\)
\(=\frac{5}{23}.\left(\frac{17}{26}+\frac{9}{26}\right)=\frac{5}{23}.1=\frac{5}{23}\)
c) \(\left(\frac{3}{29}-\frac{1}{5}\right).\frac{29}{3}=\frac{3}{29}.\frac{29}{3}-\frac{1}{5}.\frac{29}{3}\)
\(=1-1\frac{14}{15}=\frac{14}{15}\)
Bài 3:
a) x/5 = 2/5
=> x =2
b) -4/x = 20/14 = 10/7
=> -4/x = 10/7
=> x.10 = (-4).7
x.10 = - 28
x= -28 :10
x= -2,8
c) 4/7 = 12/x = 12/ 21
=> 12/x = 12/21
=> x = 21
d) 3/7 = x / 21 = 9/21
=> x/21 = 9/21
=> x= 9
1. Tìm số nguyên x biết
a.3x+27=9
\(\Leftrightarrow3x=-18\)
\(\Leftrightarrow x=-6\)
b)\(2x^2-1=49\)
\(\Leftrightarrow2x^2=50\)
\(\Leftrightarrow2x^2=2.5^2\)
\(\Leftrightarrow x=\pm5\)
c)2x+12=3(x-7)
\(\Leftrightarrow2x+12=3x-21\)
\(\Leftrightarrow33=x\) hay \(x=33\)
d)\(\left|-9-x\right|-5=12\)
\(\Leftrightarrow\left|-9-x\right|=17\)
\(\Leftrightarrow\left[{}\begin{matrix}-9-x=17\\-9-x=-17\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-26\\x=8\end{matrix}\right.\)
đ)(x-5)(x+6)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-6\end{matrix}\right.\)
e)(3-x)(x+7)=0
\(\Leftrightarrow\left[{}\begin{matrix}3-x=0\\x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\)
2.cho biểu thức
A=(-a-b+c)-(-a-b-c)
a)rút gọn
\(A=\left(-a+b+c\right)-\left(-a-b-c\right)\)
\(A=-a+b+c+a+b+c\)
\(A=2c\)
b)Tính giá trị của A khi a=1;b=-1;c=-2
Thay a=1;b=-1;c=-2 vào A ta có
\(A=2c\)
\(A=2.\left(-2\right)\)
\(A=-4\)