1. tìm x thuộc N biết:
a) 120 - 5 . ( x + 2 ) = 45
b) ( 2.x - 3 )2 = 49
c) 5x - 1 = 24
d) [( 6.x - 72) : 2 - 84] . 28 =5628
2. a) Tìm x biết :
{ x2 - [ 82 - ( 52 - 8.3 )3-7.9 ]3 4.12 } = 1
b) Cho A = 9 ; 92 ; 93 ;...; 9100
Tìm số tự nhiên n để 8A + 9 = 9 n+3
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\(30-\left[4\left(x-2\right)+15\right]=3\) \(\left(8x-120:4\right).3^3=3^6\)
\(4\left(x-2\right)+15=27\) \(\left(8x-120:4\right)=27\)
\(4\left(x-2\right)=12\) \(8x-30=27\)
\(x-2=3\) \(8x=57\)
\(x=5\) \(x=\frac{57}{8}\)
Lời giải:
\(\left\{x^2-[8^2-(5^2-8.3)^3-7.9]^3-4.12\right\}^3=1\\ \Rightarrow x^2-[8^2-(5^2-8.3)^3-7.9]^3-4.12=1\\ \Rightarrow x^2-(64-1-63)^3-48=1\\ \Rightarrow x^2-48=1\\ \Rightarrow x^2=49=7^2=(-7)^2\\ \Rightarrow x=\pm 7\)
2a/ 2x - 3 = 16 => 2x - 3 = 24 => x - 3 = 4 => x = 7
b/ {x2 - [82 - (52 - 8.3)3 - 7.9]3 - 4.12}3 = 1
=> x2 - [82 - (52 - 8.3)3 - 7.9]3 - 4.12 = 1
=> x2 - [64 - (25 - 8.3)3 - 7.9]3 = 1 + 4.12 = 49
=> x2 - (64 - 13 - 63)3 = 49
=> x2 - 0 = 49
=> x2 = 49
=> x = 7
a) Ta có: \(x^2-2x+1=25\)
\(\Leftrightarrow\left(x-1\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
b) Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)
\(\Leftrightarrow10x=20\)
hay x=2
c) Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^3-1-x\left(x^2-4\right)=5\)
\(\Leftrightarrow x^3-1-x^3+4x=5\)
\(\Leftrightarrow4x=6\)
hay \(x=\dfrac{3}{2}\)
d) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x=-10\)
hay \(x=-\dfrac{5}{12}\)
a,\(< =>\left(x-1\right)^2-5^2=0< =>\left(x-1-5\right)\left(x-1+5\right)=0\)
\(< =>\left(x-6\right)\left(x+4\right)=0=>\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
b,\(< =>25x^2+10x+1-25x^2+9-30=0\)
\(< =>10x-20=0< =>10\left(x-2\right)=0< =>x=2\)
c,\(< =>x^3-1-x\left(x^2-4\right)-5=0\)
\(< =>x^3-1-x^2+4x-5=0< =>4x-6=0< =>x=\dfrac{6}{4}\)\(d,< =>\left(x-2\right)^3-x^3+3^3+6x^2+12x+6-15=0\)
\(< =>x^3-6x^2+12x-x^3+6x^2+12x+10=0\)
\(< =>24x+10=0< =>x=-\dfrac{5}{12}\)
a: Ta có: \(x^2-2x+1=25\)
\(\Leftrightarrow\left(x-4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=6\end{matrix}\right.\)
b: Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)
\(\Leftrightarrow10x=20\)
hay x=2
c: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^3-1-x\left(x^2-4\right)=5\)
\(\Leftrightarrow x^3-1-x^3+4x=5\)
\(\Leftrightarrow4x=6\)
hay \(x=\dfrac{3}{2}\)
\(1,\\ a,=-35x^5y^4z\\ b,=6x^2-30x-6x^2-3x=-33x\\ c,=x^3-9x^2-2x^2+18x-x+9=x^3-11x^2+17x+9\\ 2,\\ A\left(x\right)+B\left(x\right)=10-2x+4x^3-5x^2-10x^3-5x+6x^2-20\\ =-6x^3+x^2-7x-10\\ A\left(x\right)-B\left(x\right)=10-2x+4x^3-5x^2+10x^3+5x-6x^2+20\\ =14x^3-11x^2+3x+30\\ 3,\\ a,M\left(x\right)=5x+20=0\\ \Leftrightarrow x=-4\\ b,N\left(x\right)=100x^2-49=0\\ \Leftrightarrow\left(10x-7\right)\left(10x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\\ c,P\left(x\right)=3x-15=0\\ \Leftrightarrow x=5\)
Bài 1;
a)\(5x^3yz.\left(-7x^2y^3\right)=-35.x^5y^4z\)
b)\(6x\left(x-5\right)-x\left(6x+3\right)=6x^2-30x-6x^2-3x=-33x\)
c) \(\left(x-9\right)\left(x^2-2x-1\right)=x^3-2x^2-x-9x^2+18x+9=x^3-11x^2+17x+9\)
a) 120 - 5 . ( x + 2 ) = 45
5 . (x + 2) = 120 - 45
5 . (x + 2) = 75
x + 2 = 75 : 5
x + 2 = 15
x = 17
b) ( 2.x - 3 )2 = 49
( 2.x - 3 )2 = 72
( 2.x - 3 ) = 7
2x = 10
x = 5