Tìm số nguyên \(x\), biết :
\(11-\left(15+11\right)=x-\left(25-9\right)\)
Tìm số nguyên x:
a) \(11-\left(15-49\right)=-x+(9-25)\)
b) \(x-+\left(-479-479\right)=-\left(26-74\right)\)
c) \(-\left(1930+2020\right)-x=\left(-96\right)+\left(-134\right)\)
\(11-\left(15-49\right)=-x+\left(9-25\right)\)
\(11-15+49=-x+9-25\)
\(x=9-25-11+15-49\)
\(x=-61\)
Vậy \(x=-61\)
\(x-+\left(-479-479\right)=-\left(26-74\right)\)
\(x+479+479=-26+74\)
\(x=74-26-479-479\)
\(x=-910\)
Vậy \(x=-910\)
\(-\left(1930+2020\right)-x=\left(-96\right)+\left(-134\right)\)
\(x=-3950+96+134\)
\(x=-3722\)
Vậy \(x=-3722\)
BT9: Tìm x biết
\(9,\left(2x-5\right)^2-\left(x+1\right)^2=0\)
\(10,\left(x+3\right)^2-x^2=45\)
\(11,\left(5x-4\right)^2-49x^2=0\)
\(12,16\left(x-1\right)^2-25=0\)
\(9,\left(2x-5\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(2x-5-x-1\right)\left(2x-5+x+1\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\3x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(S=\left\{6;\dfrac{4}{3}\right\}\)
\(10,\left(x+3\right)^2-x^2=45\)
\(\Leftrightarrow x^2+6x+9-x^2-45=0\\ \Leftrightarrow6x=36\\ \Leftrightarrow x=6\)
Vậy \(S=\left\{6\right\}\)
\(11,\left(5x-4\right)^2-49x^2=0\\ \Leftrightarrow\left(5x-4\right)^2-\left(7x\right)^2=0\\ \Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\\ \Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\12x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(S=\left\{-2;\dfrac{1}{3}\right\}\)
\(12,16\left(x-1\right)^2-25=0\\ \Leftrightarrow4^2\left(x-1\right)^2-5^2=0\\ \Leftrightarrow\left[4\left(x-1\right)\right]^2-5^2=0\\ \Leftrightarrow\left(4x-4\right)^2-5^2=0\\ \Leftrightarrow\left(4x-4-5\right)\left(4x-4+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-9=0\\4x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{1}{4};\dfrac{9}{4}\right\}\)
tìm x,biết:
a)\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
b)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
c)\(\left(x+2\right)^2=\frac{38}{25}+\frac{9}{10}-\frac{11}{15}+\frac{13}{21}-\frac{15}{28}+\frac{17}{36}-...+\frac{197}{4851}-\frac{199}{4950}\)
giúp tớ với,huhu
Tìm x:
\(\frac{\left(13\frac{2}{9}-15\frac{2}{3}\right)\cdot\left(30^2-5^4\right)}{\left(18\frac{3}{7}-17\frac{1}{4}\right)\cdot\left(25-12\cdot5^2\right)}\cdot x=\frac{\frac{2}{11}+\frac{3}{13}+\frac{4}{15}+\frac{5}{17}}{4\frac{1}{11}+\frac{5}{13}+\frac{9}{15}+\frac{13}{17}}\)
Chứng minh rằng với mọi số nguyên \(x\) thì biểu thức \(P\) là một số chính phương. \(P=\left(x+5\right)\left(x+7\right)\left(x+9\right)\left(x+11\right)+16\).
\(p=\left[\left(x+5\right).\left(x+11\right)\right].\left[\left(x+7\right).\left(x+9\right)\right]+16=\)
\(=\left(x^2+16x+55\right)\left(x^2+16x+63\right)+16=\)
\(=\left(x^2+16x\right)^2+118.\left(x^2+16x\right)+3481=\)
\(=\left(x^2+16x\right)^2+2.\left(x^2+16x\right).59+59^2=\)
\(=\left[\left(x^2+16x\right)+59\right]^2\) là một số chính phương
Tìm số nguyên x biết \(\left(x^2-\frac{4}{25}\right)\left(x^2-4\right)\left(x^2-\frac{16}{9}\right)\left(x^2-10\right)< 10\)
so sánh: \(A=26^2-24^2\) và \(B=27^2-25^2\)
tìm x, biết:
\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)
Bài 1:
\(A=26^2-24^2=\left(26-24\right)\left(26+24\right)=2\cdot50=100\)
\(B=27^2-25^2=\left(27-25\right)\left(27+25\right)=2\cdot52=104\)
=>A<B
Bài 2:
\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)
=>\(4\left(x^2+2x+1\right)+4x^2-4x+1-8\left(x^2-1\right)=11\)
=>\(4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)
=>4x+13=11
=>4x=-2
=>\(x=-\dfrac{1}{2}\)
Tìm x \(\left(x-2\right)^3-\left(x+5\right)\left(x^2-5x+25\right)+6x^2=11\)
\(\left(x-2\right)^3-\left(x+5\right)\left(x^2-5x+25\right)+6x^2=11\)
=>\(x^3-6x^2+12x-8-\left(x^3+125\right)+6x^2=11\)
=>\(x^3+12x-8-x^3-125=11\)
=>12x-133=11
=>12x=144
=>\(x=\dfrac{144}{12}=12\)
Tìm số nguyên x biết :
\(\left(7x-11\right)^3=\left(-3\right)^2\cdot15+208\)
( 7x - 11 ) 3 = ( -3 )2 .15 + 208
( 7x - 11 ) 3 = 9 .15 + 208
( 7x - 11 ) 3 = 135 + 208
( 7x - 11 ) 3 = 343
( 7x - 11 ) 3 = 73
=> 7x - 11 = 7
=> 7x = 7 + 11 = 18
=> x = 18/7
\(\left(7x-11\right)^3=\left(-3\right)^2.15+208\)
\(\Leftrightarrow\)\(\left(7x-11\right)^3=9.15+208\)
\(\Leftrightarrow\)\(\left(7x-11\right)^3=135+208\)
\(\Leftrightarrow\)\(\left(7x-11\right)^3=343\)
\(\Leftrightarrow\)\(\left(7x-11\right)^3=7^3\)
\(\Leftrightarrow\)\(7x-11=7\)
\(\Leftrightarrow\)\(7x=18\)
\(\Leftrightarrow\)\(x=\frac{18}{7}\)
Vậy \(x=\frac{18}{7}\)
Chúc bạn học tốt ~