a) 98 + 87 + 76 + ... + 32 + 21 - 12 - 23 - ... - 78 - 89

= ( 98 - 89 ) + ( 87 - 78 ) + ... + ( 32 - 23 ) + ( 21 - 12 )

= 9 + 9 + 9 + ... + 9 ( ? số 9 )

Từ 21 đến 98 với khoảng cách 11 có số các số hạng là : ( 98 - 21 ) / 11 + 1 = 8 ( số hạng )

-> 9 * 8 = 72

b) ( 66 - 1 ) * ( 66 - 2 ) * ... * ( 66 - 98 ) * ( 66 - 99 )

Trong tích trên có thừa số 66 - 66 với giá trị bằng 0.

Vậy tích trên bằng 0.

c) cái này phức tạp lắm, rất khó mà tính

câu hỏi này dễ thật

Bài 4:

a: xy=-2

=>\(x\cdot y=1\cdot\left(-2\right)=\left(-2\right)\cdot1=\left(-1\right)\cdot2=2\cdot\left(-1\right)\)

=>\(\left(x,y\right)\in\left\{\left(1;-2\right);\left(-2;1\right);\left(-1;2\right);\left(2;-1\right)\right\}\)

b: \(\left(x-1\right)\left(y+2\right)=-3\)

=>\(\left(x-1\right)\cdot\left(y+2\right)=1\cdot\left(-3\right)=\left(-3\right)\cdot1=-1\cdot3=3\cdot\left(-1\right)\)

=>\(\left(x-1;y+2\right)\in\left\{\left(1;-3\right);\left(-3;1\right);\left(-1;3\right);\left(3;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(2;-5\right);\left(-2;-1\right);\left(0;1\right);\left(4;-3\right)\right\}\)

Bài 3:

a: \(x\left(x+9\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x+9=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\)

b: \(\left(x-5\right)^2=9\)

=>\(\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=3+5=8\\x=-3+5=2\end{matrix}\right.\)

c: \(\left(7-x\right)^2=-64\)

mà \(\left(7-x\right)^2>=0\forall x\)

nên \(x\in\varnothing\)

Bài 2:

a: \(\left(-31\right)\cdot x=-93\)

=>\(31\cdot x=93\)

=>\(x=\dfrac{93}{31}=3\)

b: \(\left(-4\right)\cdot x=-20\)

=>\(4\cdot x=20\)

=>\(x=\dfrac{20}{4}=5\)

c: \(5x+1=-4\)

=>\(5x=-4-1=-5\)

=>\(x=-\dfrac{5}{5}=-1\)

d: \(-12x+1=-4\)

=>\(-12x=-4-1=-5\)

=>\(12x=5\)

=>\(x=\dfrac{5}{12}\)

`@` `\text {Ans}`

`\downarrow`

`a)`

`(2x - 1)^2 + 1 = 26`

`\Rightarrow (2x - 1)^2 = 26 - 1`

`\Rightarrow (2x - 1)^2 = 25`

`\Rightarrow (2x - 1)^2 = (+-5)^2`

`\Rightarrow`\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)

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

`\Rightarrow`\(\left[{}\begin{matrix}x=6\div2\\x=-4\div2\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy, `x \in`\(\left\{-2;3\right\}\)

`b)`

`(2x - 4)^3 + 2 = 66`

`\Rightarrow (2x - 4)^3 = 66 - 2`

`\Rightarrow (2x - 4)^3 = 64`

`\Rightarrow (2x - 4)^3 = 4^3`

`\Rightarrow 2x - 4 = 4`

`\Rightarrow 2x = 8`

`\Rightarrow x = 8 \div 2`

`\Rightarrow x = 3`

Vậy, `x = 3`

`c)`

\(7^{x+2}+5\cdot7^{x+1}+15=603\)

`\Rightarrow 7^x . 7^2 + 5 . 7^x . 7 = 603 - 15`

`\Rightarrow 7^x . 7^2 + 35 . 7^x = 588`

`\Rightarrow 7^x . (7^2 + 35) = 588`

`\Rightarrow 7^x . 84 = 588`

`\Rightarrow 7^x = 588 \div 84`

`\Rightarrow 7^x = 7`

`\Rightarrow 7^x = 7^1`

`\Rightarrow x = 1`

Vậy, `x = 1.`

\(#48Cd\)

\(a)\left(2x-1\right)^2+1=26\)

\(\left(2x-1\right)^2=25\)

\(TH1:2x-1=5\)

\(2x=6\)

\(x=3\)

\(TH2:2x-1=-5\)

\(2x=-4\)

\(x=-2\)

Vậy........

\(b)\left(2x-4\right)^3+2=66\)

\(\left(2x-4\right)^3=64=4^3\)

\(2x-4=4\)

\(2x=8\)

\(x=4\)

\(c)7^x+2+5.7^x+1+15=603\)

\(7^x\left(1+5\right)=603-15-1-2\)

\(7^x.6=585\)

Bạn xem lại phần này nhé . x tìm ra không được chẵn lắm á cậu.

Bạn xem lại câu a

a,(2x-1)²+1=26

=>(2x-1)²=26-1=25=5²

=>2x-1=5 và -5

=>x=(5+1):2=3 và x=(-5+1):2=-2

b,(2x-4)³+2=66

=>(2x-4)³=66-2=64=4³

=>2x-4=4

=>x=(4+4):2=4

c,7^x+2+5\(\cdot\)7^x+1+15=603

=>7^x\(\cdot\)7²+5\(\cdot\)7^x\(\cdot\)7=603-15=588

=>7^x(7²+5\(\cdot\)7)=588

=>7^x\(\cdot\)84=588

=>7^x=588:84=7

=>x=1

a) 1 267 + 99 + 501 = 1 267 + (99 + 501)

                                = 1 267 + 600 = 1 867

     25 x 14 x 4 = (25 x 4) x 14

                       = 100 x 14 = 1 400

b) 3 905 x (50 – 1) = 3 905 x 50 – 3 905 x 1

                             = 195 250 – 3 905 = 191 345

     270 : ( 27 x 2 ) = 270 : 27 : 2

                             = 10 : 2 = 5

c) 115 x 58 + 115 x 42 = 115 x (58 + 42)

                                   = 115 x 100 = 11 500

136 x 67 – 136 x 66 = 136 x (67 – 66)

                              = 136 x 1 = 136

4)

a,

\(34^2+66^2+68\cdot66\\ =34^2+68\cdot66+66^2\\=34^2+2\cdot34\cdot68+66^2\\ =\left(34+66\right)^2\\ =100^2 =10000\)

b,

\(74^2+24^2-48\cdot74\\ =74^2-48\cdot74+24^2\\ =74^2-2\cdot24\cdot74+24^2\\ =\left(74-24\right)^2\\ =50^2=2500\)

c,

\(729^2-728^2\\ =\left(729+728\right)\left(729-728\right)\\ =1457\cdot1\\ =1457\)

d,

\(1001^2-1\\ =1001^2-1^2\\ =\left(1001+1\right)\left(1001-1\right)\\ =1002\cdot1000\\ =1002000\)

5)

a,

\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ =1\cdot\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ =\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ =\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\\ =\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\\ =\left(2^8-1\right)\left(2^8+1\right)\\ =2^{16}-1\)

b,

\(7\cdot\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\\ =\left(2^3-1\right)\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\\ =\left(2^6-1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\\ =\left(2^{12}-1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\\ =\left(2^{24}-1\right)\left(2^{24}+1\right)\\ =2^{48}-1\)

\(a,\frac{3}{4}-\frac{3}{4}:x=\frac{1}{2}\)

\(\frac{3}{4}:x=\frac{3}{4}-\frac{1}{2}\)

\(\frac{3}{4}:x=\frac{1}{4}\)

\(x=\frac{3}{4}:\frac{1}{4}\)

\(x=3\)

1. Tính:

a) 23 + 24 - ( 65 x 2 - 4 ) x 4 + 66 + 44 - ( 54 : 9 x 9 : 9 + 45 ) : 1 + 1 - 1 x ( 23 + 23 - 23 x 2 +1 ) = 51

b) 4444444444444444444444444444444444444444444444444444444444444444444 : 1

= 4444444444444444444444444444444444444444444444444444444444444444444

c) 3⁰ = 0

d) 1/1 = 1

e)

a) - 530

b) 44444444444444444444444444444444444444444444444444444444444444444444

c) 0

d) 1

e) Em là em

-530

444444444444444444444444444444444444444444444444444444444444444

0

1

...

a) \(\frac{3}{4}-\frac{3}{4}:x=\frac{1}{2}\)

\(\Leftrightarrow\frac{3}{4}:x=\frac{1}{4}\Leftrightarrow x=3\)

b) \(-12.\left(x-5\right)+7.\left(3-x\right)=5\)

\(\Leftrightarrow-12x+60+21-7x=5\Leftrightarrow-19x+81=5\Leftrightarrow-19x=-76\)

\(\Leftrightarrow x=4\)

c) \(\left|7x+3\right|=66\Leftrightarrow7x+3=66\text{ hoặc }-7x-3=66\)

\(\Leftrightarrow7x=63\text{ hoặc }-7x=69\)

\(\Leftrightarrow x=9\text{ hoặc }x=-\frac{69}{7}\)

d) \(\left(x-7\right).\left(x+3\right)< 0\Leftrightarrow x-7< 0\text{ và }x+3>0\Leftrightarrow x< 7\text{ và }x>-3\)

\(\Leftrightarrow-3< x< 7\)