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![](https://rs.olm.vn/images/avt/0.png?1311)
A= (3,1 - 2,5) - (-2,5 + 3,1)
=3.1-2.5+2.5-3.1
=0
B= (5,3 - 2,8) - (4 + 5,3)
=5.3 - 2.8-4-5.3
= -6.8
C= -(251.3 + 281) + 3.251 - (1-281)
=-251.3-281+3.251-1+281
=-1
D= -(3/5+3/4)−(−3/4+2/5)
= -3/5-3/4+3/4-2/5
= -5/5
=-1
\(A=\left(3,1-2,5\right)-\left(-2,5+3,1\right)=3,1-2,5+2,5-3,1=0\)
\(B=\left(5,3-2,8\right)-\left(4+5,3\right)=5,3-2,8-4-5,3=-6,8\)
\(C=-\left(215\cdot3+281\right)+3\cdot215-\left(1-281\right)=-215\cdot3-281+3\cdot215-1+281=1\)
\(D=-\left(\frac{3}{5}+\frac{3}{4}\right)-\left(-\frac{3}{4}+\frac{2}{5}\right)=-\frac{3}{5}-\frac{3}{4}+\frac{3}{4}-\frac{2}{5}=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = \(\left(3,1-2,5\right)-\left(-2,5+3,1\right)\)
\(=3,1-2,5+2,5-3,1=6,2\)
B = \(\left(5,3-2,8\right)-\left(4+5,3\right)\)
\(=5,3-2,8-4-5,3=-6,8\)
C = \(-\left(251.3+281\right)+3.251-\left(1-281\right)\)
\(=-251.3-281+3.251-1+281=-1\)
D = \(-\left(\frac{3}{5}+\frac{3}{4}\right)-\left(-\frac{3}{4}+\frac{2}{5}\right)\)
\(=-\frac{3}{5}-\frac{3}{4}+\frac{3}{4}-\frac{2}{5}=-1\)
(Nhớ k cho mình với nhá!)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(x+1\right)^2=81\)
\(\left(x+1\right)^2=9^2\) Hoặc \(\left(x+1\right)^2=\left(-9\right)^2\)
\(\left(x+1\right)=9\) \(x+1=-9\)
\(x=8\) \(x=-10\)
b,\(\left(x+5\right)^{^{ }3}=-64\)
\(\left(x+5\right)^3=\left(-4\right)^3\)
\(x+5=-4\)
=> \(x=-9\)
c,\(\left(2x-3\right)^2=9\)
=>\(\left(2x-3\right)^2=3^2\)Hoặc \(\left(2x-3\right)^2=\left(-3\right)^2\)
\(2x-3=3\) \(2x-3=-3\)
\(2x=6\) \(2x=0\)
=> \(\hept{\begin{cases}x=3\\x=0\end{cases}}\)
d, \(\left(4x+1\right)^3=27\)
\(\left(4x+1\right)^{^{ }3}=3^3\)
\(4x+1=3\)
\(4x=2\)
\(x=\frac{1}{2}\)
\(D=\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{8^6}{4}=\frac{\left(2^3\right)^6}{2^2}=\frac{2^{18}}{2^2}=2^{16}\)
\(D=\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{4^{15}+4^{10}}{4^6+4^{11}}=\frac{4^{10}.4^5+4^{10}}{4^6+4^6.4^5}=\frac{4^{10}.\left(4^5+1\right)}{4^6.\left(4^5+1\right)}=\frac{4^{10}}{4^6}=4^4=256\)
phần D trên mk làm sai xin lỗi nha
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có:
\(\frac{4}{15}+\frac{1}{6}-\frac{4}{9}>\frac{2}{3}-x-\frac{1}{4}\\ \Rightarrow x+\frac{4}{15}+\frac{1}{6}-\frac{4}{9}>\frac{2}{3}-\frac{1}{4}\\ \Rightarrow x>\frac{2}{3}+\frac{4}{9}-\frac{1}{4}-\frac{1}{6}-\frac{4}{15}\\ \Rightarrow x>\left(\frac{6}{9}+\frac{4}{9}\right)-\left(\frac{15}{60}+\frac{10}{60}+\frac{16}{60}\right)\)
\(x>\frac{10}{9}-\frac{41}{60}\\ x>\frac{200-123}{180}\Rightarrow x>\frac{77}{180}\)
b) Bất đẳng thức kép
\(4-1\frac{1}{3}< x+\frac{1}{5}< 12\frac{2}{7}-3\frac{3}{8}\)
có nghĩa là ta phải có hai bất đẳng thức đồng thời:
\(x+\frac{1}{5}>4-1\frac{1}{3}\) và \(x+\frac{1}{5}< 12\frac{2}{7}-3\frac{3}{8}\)
Ta tìm các giá trị của x cần thỏa mãn bất đẳng thức thứ nhất:
\(x+\frac{1}{5}>4-1\frac{1}{3}\Rightarrow x>4-1\frac{1}{3}-\frac{1}{5}\\ \Rightarrow x>\frac{37}{15}\)
Từ bất đẳng thức thứ hai
\(x+\frac{1}{5}< 12\frac{2}{7}-3\frac{3}{8}\Rightarrow x< \frac{86}{7}-\frac{27}{8}-\frac{1}{5}\\ \Rightarrow x< \frac{2439}{280}.\)
Như vậy các số hữu tỉ x cần thỏa mãn:
\(\frac{37}{15}< x< \frac{2439}{280}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
a)\(\frac{3}{4}-0,25-\left[\frac{7}{3}+\left(-\frac{9}{2}\right)\right]-\frac{5}{6}\)
\(=\frac{3}{4}-\frac{1}{4}-\frac{14}{6}+\frac{27}{6}-\frac{5}{6}\)
\(=\frac{1}{2}-\frac{4}{3}\)
\(=-\frac{5}{6}\)
b)\(7+\left(\frac{7}{12}-\frac{1}{2}+3\right)-\left(\frac{1}{12}+5\right)\)
\(=7+\frac{1}{12}+3-\frac{1}{12}-5\)
\(=5\)
Câu 2:
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
\(-\frac{1}{12}\le\frac{x}{12}< 1-\frac{5}{12}\)
\(-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)
Vậy -1\(\le\)x<7
a) -(251 x3 + 281) + 3x 251- (1-281) b) \(-\left(\frac{3}{54}+\frac{3}{4}\right)-\left(\frac{-3}{4}+\frac{2}{5}\right)\)
=-251x3 - 281+3 x 251- 1+ 281 =\(\frac{-1}{18}-\frac{3}{4}+\frac{3}{4}-\frac{2}{5}\)
=[(-251 x 3)+ (3 x 251)]+(281-281) -1 = (\(\frac{3}{4}-\frac{3}{4}\))-(\(\frac{2}{5}+\frac{1}{18}\))
=0+ 0-1 =0 -41/ 90
=\(-\frac{41}{90}\)
= -1