2.|x-1|-9=-13+17
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a) \(326-2x=78\Rightarrow2x=326-78=248\)
\(\Rightarrow x=248:2=124\)
b) \(42.5-\left(3x+6\right)=3^4\Rightarrow210-\left(3x+6\right)=81\)
\(\Rightarrow3x+6=210-81=129\Rightarrow3x=129-6=123\)
\(\Rightarrow x=123:3=41\)
a, 326 -2x =78
\(\Rightarrow\)2x=326-78=248
\(\Rightarrow\)x=248/2=124
b,\(42.5-\left(3x+6\right)=3^4\)
\(\Rightarrow\)\(210-3x-6=81\)
\(\Rightarrow\)3x=123
\(\Rightarrow\)x=123/3=41
c, \(4^x:64=4^{250}\)
\(\Rightarrow\)\(4^x:4^3=4^{250}\)
\(\Rightarrow\)x=253
d, \(30x+1+2+3+...+30=495\)
\(\Rightarrow\)\(30x+465=495\)
\(\Rightarrow\)\(x=1\)
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#chanh
Sửa : 72 thành 17 :v
TBR : a chia 17 dư 8 => a + 9 \(⋮\)17 (1)
a chia 25 dư 16 => a + 9 \(⋮\)25 (2)
Từ (1),(2) => a + 9 \(\in\)BC(17,25)
Ta có :
17 = 17
25 = 52
=> BCNN (17,25) = 17 . 52 = 425
B(425) = BC(17,25) = {0;425 ; 850; 1275;..}
Mà a có 3 chữ số và a lớn nhất
=> a + 9 lớn nhất và có 3 chữ số
=> a + 9 = 850
=> a = 841
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\frac{a}{5}=\frac{b}{6}=\frac{c}{7}=k\Rightarrow\hept{\begin{cases}a=5k\\b=6k\\c=7k\end{cases}}\)
\(\Rightarrow4\left(a-b\right)\left(b-c\right)=4\left(5k-6k\right)\left(6k-7k\right)=4.\left(-k\right).\left(-k\right)=4k^2\)(1)
và \(\left(c-a\right)^2=\left(7k-5k\right)^2=\left(2k\right)^2=4k^2\)(2)
Từ (1) và (2) suy ra \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có;
\(\frac{a}{5}=\frac{b}{6}=\frac{c}{7}=\frac{a-b}{-1}=\frac{b-c}{-1}=\frac{c-a}{2}\)
\(\Leftrightarrow\hept{\begin{cases}a-b=b-c\\c-a=-2\left(b-c\right)=-2\left(a-b\right)\end{cases}}\)
\(\left(c-a\right)^2=-2\left(a-b\right)\cdot-2\left(b-c\right)=4\left(a-b\right)\left(b-c\right)\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x.y+x+y=8\)
\(x.y+x+y+1=8+1\)
\(x\left(y+1\right)+\left(y+1\right)=9\)
\(\left(y+1\right)\left(x+1\right)=9\)
Mà \(9=3.3=1.9=\left(-3\right)\left(-3\right)=\left(-1\right)\left(-9\right)\)
Lập bảng, ta có:
\(x+1=\) | \(1\) | \(9\) | \(3\) | \(-1\) | \(-9\) | \(-3\) |
\(\Rightarrow x=\) | \(0\) | \(8\) | \(2\) | \(-2\) | \(-10\) | \(-4\) |
\(y+1=\) | \(9\) | \(1\) | \(3\) | \(-9\) | \(-1\) | \(-3\) |
\(\Rightarrow y=\) | \(8\) | \(0\) | \(2\) | \(-10\) | \(-2\) | \(-4\) |
Vậy các cặp (x;y) thỏa mãn là: \(\left(0;8\right);\left(8;0\right);\left(2;2\right);\left(-2;-10\right);\left(-10;-2\right);\left(-4;-4\right)\)
xy+x+y=8
x(y+1)+(y+1)=9
(y+1)(x+1)=9
Ta có:9=1.9=(-1).(-9)=3.3=(-3).(-3)
Kẻ bảng....
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{2.x}-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{45.46}=-2\)
\(\frac{1}{2.x}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{45.46}\right)=-2\)
\(\frac{1}{2.x}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{45}-\frac{1}{46}\right)=-2\)
\(\frac{1}{2.x}-\left(1-\frac{1}{46}\right)\)
\(\frac{1}{2.x}-\frac{45}{46}=-2\)
\(\frac{1}{2.x}=-2+\frac{45}{46}\)
\(\frac{1}{2.x}=\frac{-47}{46}\)
\(2x=\frac{46}{-47}\)
\(x=\frac{46}{-47}:2=\frac{-23}{47}\)
2 . |x-1| - 9 = -13 + 17
2 . |x-1| - 9 = 4
2 . |x-1| = 4 + 9
2 . |x-1| = 13
|x-1| = 13 : 2
|x-1| = 13/2
=> x - 1 = 13 / 2 hoặc x - 1 = -13 / 2
+) x - 1 = 13 / 2
x = 13 / 2 + 1
x = 15 / 2
+) x - 1 = -13 / 2
x = -13 / 2 + 1
x = -11 / 2
Vậy x \(\in\) { 15 / 2 ; -11 / 2 }
\(2\cdot\left|x-1\right|-9=-13+17\)
\(\Leftrightarrow2\cdot\left|x-1\right|-9=5\)
\(\Leftrightarrow2\cdot\left|x-1\right|=5+9\)
\(\Leftrightarrow2\cdot\left|x-1\right|=14\)
\(\Leftrightarrow\left|x-1\right|=\frac{14}{2}\)
\(\Leftrightarrow\left|x-1\right|=7\Leftrightarrow x-1=\pm7\)
TH1:\(x-1=17\)
\(\Leftrightarrow x=17+1=18\)
TH2:\(x-1=-17\)
\(\Leftrightarrow x=-17+1=-16\)
Vậy x=18;-16