a)13-18-(-42)-1513-18-(-42)-15

=13-18+42-1513-18+42-15

= (-18-18)+(42+42)+(-1513+13-15)

= -36+84+(-1500-15)

= -36+84+(-1515)

=  48+(-1515)

=  -1467

Bài 46:

11: Ta có: \(-4\left|x-2\right|=-8\)

\(\Leftrightarrow\left|x-2\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)

Vậy: x∈{0;4}

12: Ta có: \(5\left|x+2\right|=-10\cdot\left(-2\right)\)

\(\Leftrightarrow5\left|x+2\right|=20\)

\(\Leftrightarrow\left|x+2\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

Vậy: x∈{-6;2}

13: Ta có: \(6\left|x-2\right|=18:\left(-3\right)\)

\(\Leftrightarrow6\left|x-2\right|=-6\)(1)

Ta có: \(\left|x-2\right|\ge0\forall x\)

\(\Rightarrow6\left|x-2\right|\ge0\forall x\)(2)

Ta có: -6<0(3)

Từ (1), (2) và (3) suy ra x∈∅

Vậy: x∈∅

14: Ta có:\(-7\left|x+4\right|=21:\left(-3\right)\)

\(\Leftrightarrow-7\left|x+4\right|=-7\)

\(\Leftrightarrow\left|x+4\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=1\\x+4=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

Vậy: x∈{-5;-3}

15: Ta có: \(4\left|x+1\right|=8\left(-2\right)-8\left(-5\right)\)

\(\Leftrightarrow4\left|x+1\right|=-16-\left(-40\right)\)

\(\Leftrightarrow4\left|x+1\right|=24\)

\(\Leftrightarrow\left|x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)

Vậy: x∈{-7;5}

16: Ta có: \(3\left|x+5\right|=-9\)(4)

Ta có: |x+5|≥0∀x

⇒3|x+5|≥0∀x(5)

Ta có: -9<0(6)

Từ (4), (5) và (6) suy ra x∈∅

Vậy: x∈∅

17: Ta có: \(-8\left|x-3\right|=24-16:2\)

\(\Leftrightarrow-8\left|x-3\right|=16\)

\(\Leftrightarrow\left|x-3\right|=-2\)

mà |x-3|≥0>-2∀x

nên x∈∅

Vậy: x∈∅

18: Ta có: \(-3\left|x+6\right|=6\cdot2-9\)

\(\Leftrightarrow-3\left|x+6\right|=3\)

\(\Leftrightarrow\left|x+6\right|=-1\)

mà |x+6|≥0>-1∀x

nên x∈∅

Vậy: x∈∅

19: Ta có: \(5-\left|x+7\right|=4\)

\(\Leftrightarrow\left|x+7\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+7=-1\\x+7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-6\end{matrix}\right.\)

Vậy: x∈{-8;-6}

20: Ta có: \(12-\left|x+8\right|=10\)

\(\Leftrightarrow\left|x+8\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=2\\x+8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-10\end{matrix}\right.\)

Vậy: x∈{-10;-6}

1:

a: Vì \(\dfrac{-4}{3}=\dfrac{-4\cdot3}{3\cdot3}=\dfrac{-12}{9}=\dfrac{12}{9}\\ \Rightarrow\dfrac{-4}{3}=\dfrac{12}{9}\)

b: Vì : \(-2\cdot3=-6\\ -6\cdot8=-48\)

nên 2 p/s ko bằng nhau 

X = ( 1+10 ) * 10 / 2 = 11 * 10 / 2 = 110 / 2 = 55

X= (2+20) * 10 /2 = 110

X= (1+19)*10/2= 100

1+2+3+4+5+6+7+8+9+10= (10+1)X10 : 2 = 55

2+4+6+8+10+12+14+16+18+20 = (20+2)x10 : 2 = 110

1+3+5+7+9+11+13+15+17+19 = (19+1) x 10 : 2 = 100

x=55

x=110

x=100

a. x = 1+ 2+ 3+ 4 + 5 + 6 + 7 + 8 + 9 +10

       = (1 +10) + (2+9) + ( 3 +8 ) + (4 + 7 ) + (5 + 6)

       = 11 +11 + 11+ 11 +11

       = 11 x 5

       = 55

b. y = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20

      = (2+20) + (4 + 18) + (6 + 16) +(8 + 14) + (10 +12)

       = 22 + 22 + 22 +22 + 22

       = 22 x 5

       = 110

mai nah

2006 x [43 x 10 - 2 x 43 x 5] +100

=2006x0+100

=0+100

=100

64x4+18x4+9x8

=256+72+72

=400

44x5+18x10+20x5

=220+180+100

=500

3x4+4x6+9x2+18

=12+24+18+18

=72

2x5+5x7+9x3

=10+35+27

=72

15:5+27:5+8:5

=[15+27+8]:5

=10

99:5-26:5-14:5

=[99-26-14]:5

=11.8

Câu cuối sai đề nha mà nếu đề như vậy thì đó là toán lớp 6

Cảm ơn bạn I LOVE YOU rất nhiều mình đang vội ,cảm ơn bạn nhé!!!

Trả lời:

\(2006\times\left(43\times10-2\times43\times5\right)+100\)

\(=2006\times\left(43\times10-43\times10\right)+100\)

\(=2006\times0+100\)

\(=0+100\)

\(=100\)

\(44\times5+18\times10+20\times5\)

\(=44\times5+18\times2\times5+20\times5\)

\(=44\times5+36\times5+20\times5\)

\(=5\times\left(44+36+20\right)\)

\(=5\times100\)

\(=500\)

\(3\times4+4\times6+9\times2+18\)

\(=12+24+18+18\)

\(=12+24+36\)

\(=72\)

\(2\times5+5\times7+9\times3\)

\(=5\times\left(2+7\right)+9\times3\)

\(=5\times9+9\times3\)

\(=9\times\left(5+3\right)\)

\(=9\times8\)

\(=72\)

\(15\div5+27\div5+8\div5\)

\(=\left(15+27+8\right)\div5\)

\(=50\div5\)

\(=10\)

\(99\div5-26\div5-14\div5\)

\(=\left(99-26-14\right)\div5\)

\(=59\div5\)

\(=\frac{59}{5}\)

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