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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:
\(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{n}\right)\)
\(A=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{n-1}{n}\)
\(A=\dfrac{1}{n}\)
nhìu dữ
a)3/2
b)-1/3
c)-5/6
d)0
e)-1/2
Bài 2
a=3
b=1/2
c=-1/3
d=0
e=9
f=-2/3
Bài 2:
Ta có: \(\dfrac{x-1}{65}+\dfrac{x-3}{63}=\dfrac{x-5}{61}+\dfrac{x-7}{59}\)
\(\Leftrightarrow\left(\dfrac{x-1}{65}-1\right)+\left(\dfrac{x-3}{63}-1\right)=\left(\dfrac{x-5}{61}-1\right)+\left(\dfrac{x-7}{59}-1\right)\)
\(\Leftrightarrow\left(x-66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\)
=>x-66=0
hay x=66
(x-100)/24 + (x-98)/26 + (x-96)/28 = 3
<=> (x - 100)/24 -1 + (x-98)/26-1 (x-96)/28 -1 = 0
<=>(x-124)/24 + (x-124)/26 + (x - 124)/28 =0
<=>(x - 124) (1/24+1/26+1/28) = 0
vì 1/24+1/26+1/28 khác 0
=> x - 124 = 0
=> x = 124
2) (x-1)/65 + (x-3)/63 = (x-5)/61 + (x-7)/59
tương tự:
(x-1)/65 -1 +(x -3)/63 -1 = (x-5)/61-1 + (x-7)/59 -1
rút gọn được:
(x - 66).(1/65 + 1/63) = (x -66).(1/61 + 1/59)
(x - 66).(1/65 + 1/63 - 1/61 -1/59) = 0
=> x = 66 (lý luận tương tự câu trên) 
Bài 3 :
Vì \(\left(x-2\right)^2\ge0\forall x\)
Nên : \(A=\left(x-2\right)^2-4\ge-4\forall x\)
Vậy \(A_{min}=-4\) khi x = 2
B1: lấy máy tính mà tính thôi bạn (nhớ lm theo từng bước)
B2:
a, \(\left|x-\frac{2}{3}\right|-\frac{1}{2}=\frac{5}{6}\)
\(\left|x-\frac{2}{3}\right|=\frac{4}{3}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{3}=\frac{4}{3}\\x-\frac{2}{3}=\frac{-4}{3}\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{-2}{3}\end{cases}}}\)
b, \(\frac{\left(-2\right)^x}{512}=-32\Rightarrow\left(-2\right)^x=-16384\Rightarrow x\in\varnothing\)
B3:
Vì \(\left(x-2\right)^2\ge0\Rightarrow A=\left(x-2\right)^2-4\ge-4\)
Dấu "=" xảy ra khi x = 2
Vậy GTNN của A = -4 khi x = 2