tính giá trị của các biểu thức sau
a) x +87002 với =707228
b) 2035 nhân x với x=100
c) x: 2 với x=84560
d)564320- x với x=304110
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12 tháng 8
Xét ΔEDI có \(\widehat{EIF}\) là góc ngoài
nên \(\widehat{EIF}=\widehat{IED}+\widehat{IDE}\)
=>\(\widehat{IED}=110^0-90^0=20^0\)
EI là phân giác của góc DEF
=>\(\widehat{DEF}=2\cdot\widehat{DEI}=40^0\)
ΔDEF vuông tại D
=>\(\widehat{DEF}+\widehat{DFE}=90^0\)
=>\(\widehat{DFE}=90^0-40^0=50^0\)
12 tháng 8
Sửa đề: \(\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{2^{12}\cdot9^6+8\cdot9^5}\)
\(=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^{12}+2^3\cdot3^{10}}\)
\(=\dfrac{2^{12}\cdot3^4\left(3-1\right)}{2^3\cdot3^{10}\left(2^9\cdot3^2+1\right)}\)
\(=\dfrac{2^9}{3^6}\cdot\dfrac{2}{1028\cdot9+1}=\dfrac{2^{10}}{729\left(1028\cdot9+1\right)}\)
12 tháng 8
Bài 2:
a: \(\left|x+\dfrac{1}{5}\right|-\dfrac{1}{2}=\dfrac{9}{10}\)
=>\(\left|x+\dfrac{1}{5}\right|=\dfrac{1}{2}+\dfrac{9}{10}=\dfrac{14}{10}=\dfrac{7}{5}\)
=>\(\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{7}{5}\\x+\dfrac{1}{5}=-\dfrac{7}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=-\dfrac{8}{5}\end{matrix}\right.\)
b: \(\dfrac{5}{4}-3\left|2x+5\right|=\dfrac{3}{4}\)
=>\(3\left|2x+5\right|=\dfrac{5}{4}-\dfrac{3}{4}=\dfrac{2}{4}=\dfrac{1}{2}\)
=>\(\left|2x+5\right|=\dfrac{1}{6}\)
=>\(\left[{}\begin{matrix}2x+5=\dfrac{1}{6}\\2x+5=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{6}-5=-\dfrac{29}{6}\\2x=-\dfrac{1}{6}-5=-\dfrac{31}{6}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{29}{12}\\x=-\dfrac{31}{12}\end{matrix}\right.\)
c: \(\left(\dfrac{3}{5}x+\dfrac{1}{2}\right)^2=\dfrac{25}{16}\)
=>\(\left[{}\begin{matrix}\dfrac{3}{5}x+\dfrac{1}{2}=\dfrac{5}{4}\\\dfrac{3}{5}x+\dfrac{1}{2}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{5}x=\dfrac{5}{4}-\dfrac{1}{2}=\dfrac{3}{4}\\\dfrac{3}{5}x=-\dfrac{5}{4}-\dfrac{1}{2}=-\dfrac{7}{4}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{3}{4}:\dfrac{3}{5}=\dfrac{5}{4}\\x=-\dfrac{7}{4}:\dfrac{3}{5}=-\dfrac{7}{4}\cdot\dfrac{5}{3}=-\dfrac{35}{12}\end{matrix}\right.\)
d: \(3-\left(2x+1\right)^2=2\)
=>\(\left(2x+1\right)^2=3-2=1\)
=>\(\left[{}\begin{matrix}2x+1=1\\2x+1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Bài 1:
a: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{9}{16}-\sqrt{\dfrac{4}{81}}:\dfrac{16}{9}+\left|-0,25\right|\)
\(=\dfrac{4}{9}\cdot\dfrac{9}{16}-\dfrac{2}{9}\cdot\dfrac{9}{16}+\dfrac{1}{4}\)
\(=\dfrac{4}{16}-\dfrac{2}{16}+\dfrac{1}{4}=\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\)
b: \(\left(-2\right)^3+\dfrac{1}{2}:\dfrac{1}{8}-\sqrt{25}+\left|-8\right|\)
\(=-8+\dfrac{1}{2}\cdot8-5+8\)
=4-5=-1
c: \(\left(\dfrac{4}{3}-\dfrac{3}{2}\right)^2-2:\left|-\dfrac{1}{9}\right|+\dfrac{-5}{18}\)
\(=\left(\dfrac{8}{6}-\dfrac{9}{6}\right)^2-2:\dfrac{1}{9}-\dfrac{5}{18}\)
\(=\dfrac{1}{36}-18-\dfrac{5}{18}=\dfrac{1}{36}-\dfrac{10}{36}-18=-\dfrac{9}{36}-18\)
\(=-18-\dfrac{1}{4}=-18,25\)
d: \(\left(-\dfrac{3}{4}\right)^2:\left(-\dfrac{1}{4}\right)^2+9\left(\dfrac{1}{3}\right)^2+\left|-\dfrac{3}{2}\right|\)
\(=\left(-\dfrac{3}{4}:\dfrac{-1}{4}\right)^2+9\cdot\dfrac{1}{9}+\dfrac{3}{2}\)
\(=3^2+1+\dfrac{3}{2}=9+1+\dfrac{3}{2}=10+\dfrac{3}{2}=11,5\)
12 tháng 8
Bài 4:
a: \(216x^3+27y^3=27\left(8x^3+y^3\right)\)
\(=27\left[\left(2x\right)^3+y^3\right]\)
\(=27\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
b: \(64a^3-8=8\left(8a^3-1\right)\)
\(=8\left[\left(2a\right)^3-1^3\right]\)
\(=8\left(2a-1\right)\left(4a^2+2a+1\right)\)
c: \(x^3+8=x^3+2^3=\left(x+2\right)\left(x^2-2x+4\right)\)
d: \(27x^3-8y^3=\left(3x\right)^3-\left(2y\right)^3\)
\(=\left(3x-2y\right)\left[\left(3x\right)^2+3x\cdot2y+\left(2y\right)^2\right]\)
\(=\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)
Bài 5:
a: \(3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)\)
\(=3\left(x^2-2xy+y^2\right)-2\left(x^2+2xy+y^2\right)-\left(x^2-y^2\right)\)
\(=3x^2-6xy+3y^2-2x^2-4xy-2y^2-x^2+y^2\)
\(=2y^2-10xy\)
b: \(\left(x-y\right)^3-3\left(x-y\right)^2\cdot x+3\left(x-y\right)\cdot x^2-x^3\)
\(=\left(x-y-x\right)^3\)
\(=\left(-y\right)^3=-y^3\)
c: \(\left(3x+3\right)^3-2\left(x+1\right)^3-\left(5x-1\right)^2\)
\(=27\left(x+1\right)^3-2\left(x+1\right)^3-\left(5x-1\right)^2\)
\(=25\left(x+1\right)^3-25x^2+10x-1\)
\(=25x^3+75x^2+75x+25-25x^2+10x-1\)
\(=25x^3+50x^2+85x+24\)
d: \(\left(-2x+3\right)^3-\left(x+1\right)^3+\left(3x-1\right)^2\)
\(=\left(-2x+3-x-1\right)\left[\left(-2x+3\right)^2+\left(-2x+3\right)\left(x+1\right)+\left(x+1\right)^2\right]+\left(3x-1\right)^2\)
\(=\left(-3x+2\right)\left(4x^2-12x+9-2x^2+x+3+x^2+2x+1\right)+\left(3x-1\right)^2\)
\(=\left(-3x+2\right)\left(3x^2-9x+13\right)+\left(3x-1\right)^2\)
\(=-9x^3+27x^2-39x+6x^2-18x+26+9x^2-6x+1\)
\(=-9x^3+42x^2-63x+27\)
a: Khi x=707228 thì x+87002=707228+87002=794230
b: Khi x=100 thì 2035xX=2035x100=203500
c: Khi x=84560 thì x:2=84560:2=42280
c: Khi x=304110 thì 564320-x=564320-304110=260210