TÍNH GIÁ TRỊ BIỂU THỨC
A=2015 {390:[430-[125 25x7] ] }
TÌM X
_x-6_ =59,25
4
+ X TRỪ 6 PHẦN 4
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
VQ
14 tháng 4 2022
a) 7/5 x 3/4 : 4/5 = 21/20 : 4/5
= 21/16
Tìm x:
b) x - 3/9 = 8/7
x = 8/7 + 3/9
X = 31/21
14 tháng 8 2023
a: A=3(x^2-y^2)-2(x-y)^2
=3(x+y)(x-y)-2(x-y)^2
=(x-y)(3x+3y-2x+2y)
=(x-y)(x+5y)
=(4+4)(4-5*4)
=8*(-16)=-128
b: \(B=\left(2x-4\right)^2+2\cdot\left(2x-4\right)\left(x+1\right)+\left(x+1\right)^2\)
=(2x-4+x+1)^2
=(3x-3)^2
Khi x=-1/2 thì B=(-3/2-3)^2=(-9/2)^2=81/4
c: \(C=x^2\left(5-4\right)+y^2\left(4-6\right)+z^2\left(6+4\right)\)
=x^2-2y^2+10z^2
=6^2-2*5^2+10*4^2
=146
d: x=9 thì x+1=10
\(D=x^{2017}-x^{2016}\left(x+1\right)+x^{2015}\left(x+1\right)-...-x^2\left(x+1\right)+x\left(x+1\right)-\left(x+1\right)\)
=x^2017-x^2017+x^2016+...-x^3-x^2+x^2+x-x-1
=-1
HM
15 tháng 8 2023
a: A=3(x^2-y^2)-2(x-y)^2
=3(x+y)(x-y)-2(x-y)^2
=(x-y)(3x+3y-2x+2y)
=(x-y)(x+5y)
=(4+4)(4-5*4)
=8*(-16)=-128
NP
1
1 tháng 11 2021
\(=1-\dfrac{17^{12}\cdot2^6\cdot3^2\cdot7}{17^6\cdot\left(3^4\cdot5^2-2^2\cdot1\right)}\)
\(=1-\dfrac{2^6\cdot3^2\cdot7}{81\cdot25-4\cdot1}\)
\(=1-\dfrac{4032}{2021}=-\dfrac{2011}{2021}\)
SC
5
8 tháng 10 2023
2:
a: \(=\dfrac{1}{3}\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)=-\dfrac{1}{3}\cdot2=-\dfrac{2}{3}\)
1:
\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)
\(=-4-\dfrac{1}{4}=-\dfrac{17}{4}\)
H9
HT.Phong (9A5)
CTVHS
8 tháng 10 2023
Bài 1:
\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)
\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)
\(A=\left(7-6-5\right)-\left(\dfrac{3}{4}+\dfrac{5}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\)
\(A=-4-\dfrac{3+5-7}{4}+\dfrac{1+4-5}{3}\)
\(A=-4-\dfrac{1}{4}+\dfrac{0}{3}\)
\(A=-\dfrac{16}{4}-\dfrac{1}{4}+0\)
\(A=\dfrac{-16-1}{4}\)
\(A=-\dfrac{17}{4}\)
Bài 2:
\(\dfrac{1}{3}\cdot-\dfrac{4}{5}+\dfrac{1}{3}\cdot-\dfrac{6}{5}\)
\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{-4-6}{5}\)
\(=\dfrac{1}{3}\cdot\dfrac{-10}{5}\)
\(=\dfrac{1}{3}\cdot-2\)
\(=-\dfrac{2}{3}\)
2 tháng 12 2023
a: \(123,15-23,15\cdot4\)
\(=100+23,15-23,15\cdot4\)
\(=100-23,15\cdot3\)
\(=100-69,45=30,55\)
b: \(18,64-8,64:\left(12,2+12,8\right)\)
\(=18,64-8,64:25\)
\(=18,64-8,64\cdot0,04\)
\(=10+8,64-8,64\cdot0,04\)
\(=10+8,64\cdot0,96=10+8,2944=18,2944\)
H
1
20 tháng 2 2022
\(A=\dfrac{5x^2+3y^2}{10x^2-3y^2}\)Thay \(\dfrac{x}{3}=\dfrac{y}{5}=k\Rightarrow x=3k;y=5k\)vào ta đc
\(A=\dfrac{5.9k^2+3.25k^2}{10.9k^2-3.25k^2}=\dfrac{120k^2}{15k^2}=8\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 8 2021
\(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)\)
\(=\left|\sqrt{3}-1\right|\left(\sqrt{3}+1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)
\(\dfrac{x-25}{\sqrt{x}-5}-\dfrac{x+4\sqrt{x}+4}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\sqrt{x}-5}-\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}+2}\)
\(=\sqrt{x}+5-\left(\sqrt{x}+2\right)=5-2=3\)
23 tháng 8 2021
a: Ta có: \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}+\sqrt{2}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
=3-1
=2
b: Ta có: \(\dfrac{x-25}{\sqrt{x}-5}-\dfrac{x+4\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\sqrt{x}+5-\sqrt{x}-2\)
=3