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a)2x-138=23.32
=>2x-138=72
=>2x =72+138
=>2x =210
=>x =210:2
=>x =105
b)20-[7(x-3)+4]=2
=>7(x-3)+4 =20-2
=>7(x-3)+4 =18
=>7(x-3) =18-4
=>7(x-3) =14
=>x-3 =14:7
=>x-3 =2
=>x =2+3
=>x =5
c)[6x-39].3.28=5628
=>[6x-39].3 =5628:28
=>[6x-39].3 =201
=>6x-39 =201:3
=>6x-39 =67
=>6x =67+39
=>6x =106
=>x =106:6
=>x =17,6
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\(a)(2367-x)-1017=205\)
\(\Leftrightarrow(2367-x)=205+1017\)
\(\Leftrightarrow(2367-x)=1222\)
\(\Leftrightarrow x=2367-1222\)
\(\Leftrightarrow x=1145\)
\(b)(6x-39):3\cdot28=5628\)
\(\Leftrightarrow(6x-39):3=\frac{5628}{28}\)
\(\Leftrightarrow(6x-39):3=201\)
\(\Leftrightarrow(6x-39)=201\cdot3\)
\(\Leftrightarrow(6x-39)=603\)
\(\Leftrightarrow6x=603+39\)
\(\Leftrightarrow6x=642\)
\(c)5x-x+2x=42\)
\(\Rightarrow5x-1x+2x=42\)
\(\Rightarrow6x=42\)
\(\Rightarrow x=7\)
d\()\)Tương tự câu a và b
e\()(x-2)(x-3)=0\)
Xét có hai trường hợp :
TH1 : x - 2 = 0 => x = 2
TH2 : x - 3 = 0 => x = 3
Vậy : \(\hept{\begin{cases}x=2\\x=3\end{cases}}\)
\(\Leftrightarrow x=\frac{642}{6}=107\)
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gọi phân số cần tìm là a/b
với a/b = -39/91 và a + b = 68
a/b = -39/91 => a = -39.b/91 thay vào a + b = 68 ta được
-39.b/91 + b =68
-39.b + 91b = 68.91
52b = 6188
b = 119 thay vào a + b = 68 ta được
a + 119 = 68
a = -51
Vậy phân số cần tìm là -51/199
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35.45 = 3.3.3.3.3.4.4.4.4.4
= (3.4).(3.4).(3.4).(3.4).(3.4)
= 12.12.12.12.12 = 125
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Gọi x là thương A:B cần tìm.Theo đề, ta có:
\(\left(\dfrac{1}{1.26}+\dfrac{1}{2.27}+...+\dfrac{1}{100.125}\right)x=\dfrac{1}{1.101}+\dfrac{1}{2.102}+...+\dfrac{1}{25.125}\)
Nhân 2 vế cho 100, ta có:
\(4\left(\dfrac{25}{1.26}+\dfrac{25}{2.27}+...+\dfrac{25}{100.125}\right)x=\dfrac{100}{1.101}+\dfrac{100}{2.102}+...+\dfrac{100}{25.125}\)
\(\Rightarrow4\left(1-\dfrac{1}{26}+\dfrac{1}{2}-\dfrac{1}{27}+...+\dfrac{1}{100}-\dfrac{1}{125}\right)x=1-\dfrac{1}{101}+\dfrac{1}{2}-\dfrac{1}{102}+...+\dfrac{1}{25}-\dfrac{1}{125}\)
\(\Rightarrow4\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-\left(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{125}\right)\right]x=\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)-\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{125}\right)\)\(\Rightarrow4x=1\Rightarrow x=\dfrac{1}{4}\)
Vậy hiệu A:B là:\(\dfrac{1}{4}\)
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\(A=2\left(1.30+2.29+3.28+...+15.16\right)=\)
\(=2.\left[1.30+2.\left(30-1\right)+3.\left(30-2\right)+...+15.\left(30-14\right)\right]=\)
\(=2.\left(1.30-1.2+2.30+3.30-2.3+4.30-3.4+...+15.30-14.15\right)=\)
\(=2.\left[30.\left(1+2+3+...+15\right)-\left(1.2+2.3+3.4+...+14.15\right)\right]=\)
Đặt \(B=1+2+3+...+15=\frac{15.\left(1+15\right)}{2}=120\)
Đặt \(C=1.2+2.3+3.4+...+14.15\)
\(\Rightarrow3C=1.2.3+2.3.3+3.4.3+...+14.15.3=\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+14.15.\left(16-13\right)=\)
\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-13.14.15+14.15.16=14.15.16\)
\(\Rightarrow C=\frac{14.15.16}{3}=5.14.16=1120\)
\(\Rightarrow A=2.\left(30.B-C\right)=2\left(30.120-1120\right)=4960\)
=4 (dư 7)