1/4

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1/4

a) 10 x 2 x 3 = 20 x 3 = 60

Giá trị của biểu thức 10 x 2 x 3 là 60.

b) 6 x 3 : 2 = 18 : 2 = 9

Giá trị của biểu thức 6 x 3 : 2 là 9.

c) 84 : 2 : 2 = 42 : 2 = 21

Giá trị của biểu thức 84 : 2 : 2 là 21.

d) 160 : 4 x 3 = 40 x 3 = 120

Giá trị của biểu thức 160 : 4 x 3 là 120.

10 x 2 x 3 = 60

6 x 3 : 2 = 9

HT tui chỉ kịp làm 2 câu đầu thui nha sorry tui fải đi ngủ đây

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

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}

Bài 1: 

1: =-5/24+16/27+3/4

=-5/24+18/24+16/27

=13/24+16/27

=117/216+128/216=245/216

2: =-1/3+1/3+6/7=6/7

3: \(=\dfrac{1}{2}-\dfrac{7}{12}+\dfrac{1}{2}=1-\dfrac{7}{12}=\dfrac{5}{12}\)

4: \(=-\dfrac{5}{8}+\dfrac{14}{25}-\dfrac{6}{10}=\dfrac{-125+112-120}{200}=\dfrac{-133}{200}\)

5:

5000000m²=5km²

42dm²=4200cm²

54m²26dm²=5426dm²

2800dm²=28m²

Bài 1:

6300 kg > 62 tạ

5 phút 20 giây < 350 giây

56 m² 8 cm² = 560008 cm²

516 năm < 6 thế kỉ

Bài 2:

a. 35 x 18 = 18 x 35

b. 23 x 7 + 3 x 23 = 23 x (7+3)

Bài 3:

a. C

b. C

c. A
 

dài quá à :(

1) x - 43 = (35 - x) - 48

=> x + x = 35 - 48 + 43

=> x + x = 30

=> x = 30 : 2

=> x = 15

2) 305 - x + 14 = 48 + (x + 23)

=> 305 - x + 14 = 48 + x + 23

=> -x - x = 48 + 23 - 14 - 305

=> -x - x = -248

=> -x = -248 : 2

=> -x = -124

=> x = 124

3) - (x - 6 + 85) = (x + 51) - 54

=> -x + 6 - 85 = x + 51 - 54

=> -x - x = 51 - 54 + 85 - 6

=> -x - x = 76

=> -x = 76 : 2

=> -x = 38

=> x = -38

4) - (35 - x - 37 - x) = 33 - x

=> -35 + x + 37 + x = 33 - x

=> x + x + x = 33 + 35 - 37

=> x + x + x = 31

=> x = 31 : 3

=> x \(=\dfrac{31}{3}\)

Vì x \(\in\) Z nên không có giá trị x nào thỏa mãn trong câu này.

5) 13 - | x | = | -4 |

=> 13 - |x| = 4

=> |x| = 13 - 4

=> |x| = 9

=> \(\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)

6) | x | - 3 + 6 = 16

=> |x| = 16 - 6 + 3

=> |x| = 13

=> \(\left[{}\begin{matrix}x=13\\x=-13\end{matrix}\right.\)

7) 35 - | 2x - 1 | = 14

=> |2x - 1| = 35 - 14

=> |2x - 1| = 21

=> \(\left[{}\begin{matrix}2x-1=21\\2x-1=-21\end{matrix}\right.=>\left[{}\begin{matrix}2x=21+1\\2x=-21+1\end{matrix}\right.=>\left[{}\begin{matrix}2x=22\\2x=-20\end{matrix}\right.=>\left[{}\begin{matrix}x=22:2\\x=-20:2\end{matrix}\right.=>\left[{}\begin{matrix}x=11\\x=-10\end{matrix}\right.\)

8) | 3x - 2 | + 5 = 9 - x

=> |3x - 2| = 9 - 5 - x

=> |3x - 2| = 4 - x

=> \(\left[{}\begin{matrix}3x-2=4-x\\3x-2=x-4\end{matrix}\right.=>\left[{}\begin{matrix}3x+x=4+2\\3x-x=-4+2\end{matrix}\right.=>\left[{}\begin{matrix}4x=6\\2x=-2\end{matrix}\right.=>\left[{}\begin{matrix}x=6:4\\x=-2:2\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{6}{4}\\x=-1\end{matrix}\right.\)

Vì x \(\in\) Z nên x = -1.

9) x - ( -25 + 7 ) > 12 - ( 15 - 14 )

=> x - (-18) > 12 - 1

=> x + 18 > 11

=> x > 11 - 18

=> x > -7

10) | 17 + ( x - 15 ) | < 4

=> \(\left[{}\begin{matrix}17+\left(x-15\right)< 4\\17+\left(x-15\right)< -4\end{matrix}\right.=>\left[{}\begin{matrix}x-15< 4-17\\x-15< -4-17\end{matrix}\right.=>\left[{}\begin{matrix}x-15< -15\\x-15< -21\end{matrix}\right.=>\left[{}\begin{matrix}x< -15+15\\x< -21+15\end{matrix}\right.=>\left[{}\begin{matrix}x< 0\\x< -6\end{matrix}\right.=>x< -6\)

11) x2 - 5x = 0

=> x . (2 - 5) = 0

=> x . (-3) = 0

=> x = 0 : (-3)

=> x = 0

12) | x-9 | . (-8) = -16

=> |x - 9| = (-16) : (-8)

=> |x - 9| = 3

=> \(\left[{}\begin{matrix}x-9=3\\x-9=-3\end{matrix}\right.=>\left[{}\begin{matrix}x=3+9\\x=-3+9\end{matrix}\right.=>\left[{}\begin{matrix}x=12\\x=6\end{matrix}\right.\)

13) | 4 - 5x | = 24 với x < hoặc = 0

=> \(\left[{}\begin{matrix}4-5x=24\\4-5x=-24\end{matrix}\right.=>\left[{}\begin{matrix}5x=4-24\\5x=4-\left(-24\right)\end{matrix}\right.=>\left[{}\begin{matrix}5x=-20\\5x=28\end{matrix}\right.=>\left[{}\begin{matrix}x=-20:5\\x=28:5\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\x=\dfrac{28}{5}\end{matrix}\right.\)

Vì x \(\le\) 0 nên x = -4

14) x . ( x - 2 ) > 0

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

15) x . ( x - 2 ) < 0

=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x-2< 0\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x< 2\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}2>x< 0\left(loại\right)\\0< x< 2\left(chọn\right)\end{matrix}\right.=>0< x< 2\)

16) (x-1) . (y+1) = 5

=> \(\left[{}\begin{matrix}x-1=5\\y+1=1\end{matrix}\right.=>\left[{}\begin{matrix}x=5+1\\y=1-1\end{matrix}\right.=>\left[{}\begin{matrix}x=6\\y=0\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x-1=1\\y+1=5\end{matrix}\right.=>\left[{}\begin{matrix}x=1+1\\y=5-1\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x-1=-1\\y+1=-5\end{matrix}\right.=>\left[{}\begin{matrix}x=-1+1\\y=-5-1\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\y=-6\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x-1=-5\\y+1=-1\end{matrix}\right.=>\left[{}\begin{matrix}x=-5+1\\y=-1-1\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\y=-2\end{matrix}\right.\)

17) x . ( y +2 ) = -8

=> \(\left[{}\begin{matrix}x=1\\y+2=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\y=-8-2\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\y=-10\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x=-1\\y+2=8\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\y=8-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x=-8\\y+2=1\end{matrix}\right.=>\left[{}\begin{matrix}x=-8\\y=1-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-8\\y=-1\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x=8\\y+2=-1\end{matrix}\right.=>\left[{}\begin{matrix}x=8\\y=-1-2\end{matrix}\right.=>\left[{}\begin{matrix}x=8\\y=-3\end{matrix}\right.\)

hoặc

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

hoặc

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

hoặc

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

hoặc

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

18) xy - 2x - 2y = 0

=> x . (y - 2) - 2y = 0

=> x . (y - 2) - 2y - 4 = -4

=> x . (y - 2) - 2 . (y - 2) = -4

=> (y - 2) . (x - 2) = -4

=> \(\left[{}\begin{matrix}y-2=1\\x-2=-4\end{matrix}\right.=>\left[{}\begin{matrix}y=1+2\\x=-4+2\end{matrix}\right.=>\left[{}\begin{matrix}y=3\\x=-2\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}y-2=-1\\x-2=4\end{matrix}\right.=>\left[{}\begin{matrix}y=-1+2\\x=4+2\end{matrix}\right.=>\left[{}\begin{matrix}y=1\\x=6\end{matrix}\right.\)

hoặc

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

hoặc

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

19) 2x - 5 \(⋮\) x - 1

=> (2x - 2) - (5 - 2) \(⋮\) x - 1

=> 2(x - 1) - 3 \(⋮\) x - 1

Vì 2(x - 1) \(⋮\) x - 1 nên 3 \(⋮\) x - 1

=> x - 1 \(\in\) Ư(3) = {-3; -1; 1; 3}

=> x \(\in\) {-2; 0; 2; 4}

P/s: Mình không bảo đảm là đúng hết nên câu nào sai thì bạn thông cảm nha~