1)\(\frac{252}{x}=\frac{84}{97}\Rightarrow\)\(\frac{84}{97}=\frac{252}{291}\Rightarrow x=291\)

6) \(\frac{y}{15}=\frac{2}{5}\Rightarrow\frac{2}{5}=\frac{6}{15}\Rightarrow x=6\)

\(\frac{x+29}{31}+\frac{x+27}{33}=\frac{x+17}{43}+\frac{x+15}{45}\)

\(\frac{x+29}{31}+1+\frac{x+27}{33}+1=\frac{x+17}{43}+1+\frac{x+15}{45}+1\)

\(\frac{x+60}{31}+\frac{x+60}{33}=\frac{x+60}{43}+\frac{x+60}{45}\)

\(\left(x+60\right)\left(\frac{1}{31}+\frac{1}{33}-\frac{1}{43}-\frac{1}{45}\right)=0\)

VÌ \(\frac{1}{31}+\frac{1}{33}-\frac{1}{43}-\frac{1}{45}\ne0\)

\(\Rightarrow x+60=0\)

\(\Rightarrow x=-60\)

Câu hỏi của honoka sonoka - Toán lớp 6 - Học toán với OnlineMath

Ai lm hộ mk nha

Nhiều quá bạn ơi

nhầm:mux3=mũ3

là số nguyên tố 

Bạn nào giúp mình, mình lại 3 cái!

Mình 3 cái

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~

1: \(=75\left(27+25-2\right)=75\cdot50=3750\)

2: \(=15\left(23+37\right)+55=15\cdot60+55=955\)

3: \(=36\cdot14+36\cdot17+36\cdot69\)

\(=36\cdot100=3600\)

4: \(=200\cdot\left(32+68\right)=200\cdot100=20000\)

b) Đặt x² + x + 1 = t > 0 (dễ c/m t > 0 rồi ha)

Khi đó, pt tương đương: \(t\left(t+1\right)=12\Leftrightarrow t^2+t-12=0\Leftrightarrow\left[{}\begin{matrix}t=3\\t=-4\left(L\right)\end{matrix}\right.\)

t = 3 suy ra \(x^2+x+1=3\Leftrightarrow x^2+x-2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Vậy...

c) Chị xem lại đề giúp em ạ.