66 + 4 x c + 4 - 1 x c
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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
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.
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
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) \(\left(x-7\right).\left(x+3\right)< 0\Leftrightarrow x-7< 0\text{ và }x+3>0\Leftrightarrow x< 7\text{ và }x>-3\)
\(\Leftrightarrow-3< x< 7\)
ai tick đến 190 thì mik tick cho cả đời
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