Rút gọn biểu thức
a) x+(-10)-[87+(-41)+(-20)-x]
b)a+(250-186)-(270-186)
c)b-(393+170)+(93+170)
d) -(a-b+c)-(a+b+c)
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Những câu hỏi liên quan
L
5 tháng 12 2019
\(a)3\left(x-1\right)^2=75\)
\(\Leftrightarrow\left(x-1\right)^2=25\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=\left(-5\right)^2\\\left(x-1\right)^2=5^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=-5\\x-1=5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-4\\x=6\end{cases}}\)
\(b)170+\left(84-5x\right):2^2=186\)
\(\Leftrightarrow\frac{84-5x}{4}=16\)
\(\Leftrightarrow84-5x=64\)
\(\Leftrightarrow5x=20\)
\(\Leftrightarrow x=4\)
\(c)125-5\left(x+4\right)=38\)
\(\Leftrightarrow5\left(x+4\right)=87\)
\(\Leftrightarrow x+4=\frac{87}{5}\)
\(\Leftrightarrow x=\frac{87}{5}-4\)
\(\Leftrightarrow x=\frac{67}{5}\)
15 tháng 7 2021
B1
a, \(=>A=\left(x+y+x-y\right)\left(x+y-x+y\right)=2x.2y=4xy\)
b, \(=>B=\left[\left(x+y\right)-\left(x-y\right)\right]^2=\left[x+y-x+y\right]^2=\left[2y\right]^2=4y^2\)
c,\(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)\)
\(=\)\(\left(x+1\right)\left(x^2-x+1\right)\left(x-1\right)\left(x^2+x+1\right)=\left(x^3+1^3\right)\left(x^3-1^3\right)=x^6-1\)
d, \(\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)
\(=\left(a+b-c\right)^2-\left(b-c\right)^2+\left(a-b+c\right)^2-\left(b-c\right)^2\)
\(=\left(a+b-c+b-c\right)\left(a+b-c-b+c\right)\)
\(+\left(a-b+c+b-c\right)\left(a-b+c-b+c\right)\)
\(=a\left(a+2b-2c\right)+a\left(a-2b\right)\)
\(=a\left(a+2b-2c+a-2b\right)=a\left(2a-2c\right)=2a^2-2ac\)
B2:
\(\)\(x+y=3=>\left(x+y\right)^2=9=>x^2+2xy+y^2=9\)
\(=>xy=\dfrac{9-\left(x^2+y^2\right)}{2}=\dfrac{9-\left(17\right)}{2}=-4\)
\(=>x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=3\left(17+4\right)=63\)
15 tháng 7 2021
Bài 1:
a) Ta có: \(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=x^2+2xy+y^2-x^2+2xy+y^2\)
=4xy
b) Ta có: \(\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x+y-x+y\right)^2\)
\(=\left(2y\right)^2=4y^2\)
c) Ta có: \(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x^3-1\right)\left(x^3+1\right)\)
\(=x^6-1\)
d) Ta có: \(\left(a+b-c\right)^2+\left(a+b+c\right)^2-2\left(b-c\right)^2\)
\(=\left(a+b-c\right)^2-\left(b-c\right)^2+\left(a+b+c\right)^2-\left(b-c\right)^2\)
\(=\left(a+b-c-b+c\right)\left(a+b-c+b-c\right)+\left(a+b+c-b+c\right)\left(a+b+c+b-c\right)\)
\(=a\cdot\left(a+2b-2c\right)+\left(a+2c\right)\left(a-2b\right)\)
\(=a^2+2ab-2ac+a^2-2ab+2ac-4bc\)
\(=2a^2-4bc\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 1
\(log_{a^3}b.log_ba=\dfrac{1}{3}.log_ab.log_ba=\dfrac{1}{3}\)
\(log_{a^{10}}b^5.log_{b^3}a^9=\dfrac{1}{10}.5.log_ab.\dfrac{1}{3}.9.log_ba=\dfrac{3}{2}\)
\(log_{a^{107}}b^{101}.log_{b^{303}}a^{428}=\dfrac{1}{107}.101.log_ab.\dfrac{1}{303}.428.log_ba=\dfrac{4}{3}.log_ab.log_ba=\dfrac{4}{3}\)
12 tháng 1
a: \(log_{a^3}b\cdot log_ba=\dfrac{1}{3}\cdot log_ab\cdot log_ba=\dfrac{1}{3}\)
b: \(log_{a^{10}}b^5\cdot log_{b^3}a^9\)
\(=\dfrac{1}{10}\cdot log_ab^5\cdot\dfrac{1}{3}\cdot log_ba^9\)
\(=\dfrac{1}{30}\cdot5\cdot log_ab\cdot9\cdot log_ba=\dfrac{45}{30}=\dfrac{3}{2}\)
c: \(log_{a^{107}}b^{101}\cdot log_{b^{303}}a^{428}\)
\(=\dfrac{1}{107}\cdot log_ab^{101}\cdot\dfrac{1}{303}\cdot log_ba^{428}\)
\(=\dfrac{1}{107}\cdot101\cdot log_ab\cdot\dfrac{1}{303}\cdot428\cdot log_ba\)
\(=4\cdot\dfrac{1}{3}=\dfrac{4}{3}\)
20 tháng 10 2021
biểu thức e viết liền quá khó phân biệt ví dụ như x +1 -\(\frac{2\sqrt{x}}{\sqrt{x-1}}\)hay là x +\(\frac{1-\sqrt{2x}}{\sqrt{x-1}}\)
17 tháng 10 2021
\(a,=x^3-16x-x^2-1-x^2+1=x^3-2x^2-16x\\ b,=y^4-81-y^4+4=-77\\ d,=a^2+b^2+c^2+2ab-2bc-2ac+a^2-2ac+c^2-2ab-2ac\\ =2a^2+b^2+2c^2-2bc-6ac\)
AH
Akai Haruma
Giáo viên
29 tháng 10 2023
Lời giải:
b.
$B=-(a-c)-(a-b+c)=-a+c-a+b-c=(-a-a)+(c-c)+b=-2a+0+b=-2a+b$
c.
$C=-(15-x)+5=-15+x+5=(-15+5)+x=-10+x$
d.
$D=[a+(a+3)]-[(a+2)-(a-2)]$
$=(2a+3)-(a+2-a+2)=(2a+3)-4=2a+3-4=2a-1$
30 tháng 1 2022
Bài 1:
\(A=\dfrac{-1}{3}+1+\dfrac{1}{3}=1\)
\(B=\dfrac{2}{15}+\dfrac{5}{9}-\dfrac{6}{9}=\dfrac{2}{15}-\dfrac{1}{9}=\dfrac{18-15}{135}=\dfrac{3}{135}=\dfrac{1}{45}\)
\(C=\dfrac{-1}{5}+\dfrac{1}{4}-\dfrac{3}{4}=\dfrac{-1}{5}-\dfrac{1}{2}=\dfrac{-7}{10}\)
Bài 2:
a: \(=\dfrac{1}{5}+\dfrac{1}{2}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{2}{21}-\dfrac{10}{21}+\dfrac{3}{20}\)
\(=\left(\dfrac{1}{5}+\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{2}{21}-\dfrac{10}{21}\right)+\left(\dfrac{1}{2}+\dfrac{3}{20}\right)\)
\(=\dfrac{-8}{21}+\dfrac{13}{20}=\dfrac{113}{420}\)
b: \(B=\dfrac{21}{23}-\dfrac{21}{23}+\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)
30 tháng 1 2022
Bài 3:
\(\dfrac{7}{3}-\dfrac{1}{2}-\left(-\dfrac{3}{70}\right)=\dfrac{7}{3}-\dfrac{1}{2}+\dfrac{3}{70}=\dfrac{490}{210}-\dfrac{105}{210}+\dfrac{9}{210}=\dfrac{394}{210}=\dfrac{197}{105}\)
\(\dfrac{5}{12}-\dfrac{3}{-16}+\dfrac{3}{4}=\dfrac{5}{12}+\dfrac{3}{16}+\dfrac{3}{4}=\dfrac{20}{48}+\dfrac{9}{48}+\dfrac{36}{48}=\dfrac{65}{48}\)
Bài 4:
\(\dfrac{3}{4}-x=1\)
\(\Rightarrow-x=1-\dfrac{3}{4}\)
\(\Rightarrow x=-\dfrac{1}{4}\)
Vậy: \(x=-\dfrac{1}{4}\)
\(x+4=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{5}-4\)
\(\Rightarrow x=-\dfrac{19}{5}\)
Vậy: \(x=-\dfrac{19}{5}\)
\(x-\dfrac{1}{5}=2\)
\(\Rightarrow x=2+\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{11}{5}\)
Vậy: \(x=\dfrac{11}{5}\)
\(x+\dfrac{5}{3}=\dfrac{1}{81}\)
\(\Rightarrow x=\dfrac{1}{81}-\dfrac{5}{3}\)
\(\Rightarrow x=-\dfrac{134}{81}\)
Vậy: \(x=-\dfrac{134}{81}\)
a) \(x+\left(-10\right)-\left[87+\left(-41\right)+\left(-20\right)-x\right]\)
\(=x-10-\left(87-41-20-x\right)=x-10-87+41+20+x\)
\(=\left(x-x\right)+\left(20-10\right)-\left(87-41\right)=10-46=-36\)
b) \(a+\left(250-186\right)-\left(270-186\right)=a+250-186-270+186\)
\(=a+\left(250-270\right)+\left(186-186\right)=a-20\)
c) \(b-\left(393+170\right)+\left(93+170\right)=b-393-170+93+170\)
\(=b+\left(170-170\right)-\left(393-93\right)=b-300\)
d) \(-\left(a-b+c\right)-\left(a+b+c\right)=-a+b-c-a-b-c=-2a-2c\)
@Vũ Cao Minh⁀ᶦᵈᵒᶫ Phèn=)
Cảm ơn nhìu lắm =)