Bài 1:

\(A=2x+2y-y\)

\(A=2x+y\)

Thay x = 2,5 và y = 3/4 vào A

\(A=2.2,5+\dfrac{3}{4}\)

\(A=5+\dfrac{3}{4}\)

\(A=\dfrac{23}{4}\)

\(B=\dfrac{5a}{3}-\dfrac{3}{b}\)

Thay a = 1/3 và b = 0,25 vào B

\(B=\dfrac{5.\dfrac{1}{3}}{3}-\dfrac{3}{0,25}\)

\(B=\dfrac{5}{9}-12\)

\(B=-\dfrac{103}{9}\)

Bài 2:

a) \(\left(2x-\dfrac{1}{2}\right).2+\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right):\dfrac{1}{8}=1\)

\(\Rightarrow4x-1+\dfrac{26}{3}=1\)

\(\Rightarrow4x+\dfrac{23}{3}=1\)

\(\Rightarrow4x=1-\dfrac{23}{3}\)

\(\Rightarrow4x=-\dfrac{20}{3}\)

\(\Rightarrow x=-\dfrac{5}{3}\)

b) \(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\)

\(\Rightarrow\dfrac{x+1}{65}+1+\dfrac{x+3}{63}+1=\dfrac{x+5}{61}+1+\dfrac{x+7}{59}+1\)

\(\Rightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}=\dfrac{x+66}{61}+\dfrac{x+66}{59}\)

\(\Rightarrow\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}\right)=\left(x+66\right)\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\)

\(\Rightarrow\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(x+66\right)\left(\dfrac{1}{61}+\dfrac{1}{59}\right)=0\)

\(\Rightarrow\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\)

Vì \(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\ne0\)

\(\Rightarrow x+66=0\)

\(\Rightarrow x=-66\)

Bài 3:

\(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{n}\right)\)

\(A=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{n-1}{n}\)

\(A=\dfrac{1}{n}\)

Câu hỏi của Thi Bùi - Toán lớp 7 - Học toán với OnlineMath

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M= |2x-3|+|x-1| với x >1,5

= 2x - 3 + x - 1 = 3x - 4

N= |2-x|-3.|x+1| với x < -1

= 2 - x + 3(x + 1) = 2 - x + 3x + 3 = 5 - 2x

m=2x-3+x-1voi x>1.5

n=2-x-3.x+1voi x<-1

Bài 7:

Cho x+5=0

 => x=-5

Cho x²-2x=0

=> x²-2x+1-1=0

=>(x-1)²-1=0

=>(x-1)²=1

=>x-1=1  thì x=2

Nếu x-1=-1 thì x=1

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