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Ta có : (-1)+3+(-5)+7+.....+[-(x-2)+x]=600
[(-1)+3]+[(-5)+7]+.....+[-(x-2)]+x=600
2 + 2 + .... + 2 = 600
2 . (1+1+ ...... + 1 ) = 600
\(\Leftrightarrow\) 1 + 1 + .... + 1 = 600 : 2
\(\Leftrightarrow\)1 + 1 + ..... + 1 = 300
Số dấu [] là : (x - 3 ) : 4 + 1
\(\Rightarrow\)(x - 3 ) : 4 + 1 = 300
\(\Rightarrow\)(x-3) : 4 = 299
\(\Rightarrow\)x - 3 = 299 x 4
\(\Rightarrow\)x - 3 = 1196
\(\Rightarrow\)x = 1196 + 3
\(\Rightarrow\)x = 1199
Vậy x = 1199.
# HOK TỐT #
a, Ta có :
xy=6
yz=-14
xz=-21
=>(xyz)2=1764=>xzy=42 hoặc -42
+)xyz=42
=>z=42:6=7
=>x=-3
=>y=-2
+)xyz=-42
=>z=-7
=>y=2
=>x=3
\(\text{Phần a :}\)
\(\frac{2}{3}.x=\frac{-4}{27}\)
\(\Rightarrow x=\frac{-4}{27}:\frac{2}{3}=\frac{-2}{9}\)
\(\text{Phần b :}\)
\(-1\frac{1}{3}.x=1\frac{1}{15}\)
\(\Rightarrow\frac{-4}{3}.x=\frac{16}{15}\)
\(\Rightarrow x=\frac{16}{15}:\frac{-4}{3}=\frac{-4}{5}\)
a) 4x3 + 12 = 120
=> 4x3 = 120 - 12
=> 4x3 = 108
=> x3 = 108 : 4
=> x3 = 27
=> x = 3
b) 3 . 2x - 3 = 45
=> 3. 2x = 45 + 3
=> 3 . 2x = 48
=> 2x = 48 :3
=> 2x = 16
=> x = 4
c) 20 - [ 7 ( x - 3 ) + 4 ] = 2
=> 7 ( x - 3 ) + 4 = 18
=> 7 ( x - 3 ) = 14
=> x - 3 = 2
=> x = 5
a) \(\left(x-5\right)-\frac{1}{3}=\frac{2}{5}\)
\(\Rightarrow\left(x-5\right)=\frac{2}{5}+\frac{1}{3}\)
\(\Rightarrow\left(x-5\right)=\frac{11}{15}\)
\(\Rightarrow x-5=\frac{11}{15}\)
\(\Rightarrow x=\frac{11}{15}+5\)
\(\Rightarrow x=\frac{86}{15}\)
b) \(\frac{2}{3}\cdot x-\frac{3}{2}\cdot x=\frac{5}{12}\)
\(\Rightarrow x\cdot\left(\frac{2}{3}-\frac{3}{2}\right)=\frac{5}{12}\)
\(\Rightarrow x\cdot\left(-\frac{5}{6}\right)=\frac{5}{12}\)
\(\Rightarrow x=\frac{5}{12}:\left(-\frac{5}{6}\right)\)
\(\Rightarrow x=-\frac{1}{2}\)
c) \(-\frac{2}{3}\cdot x+\frac{1}{5}=\frac{3}{10}\)
\(\Rightarrow-\frac{2}{3}\cdot x=\frac{3}{10}-\frac{1}{5}\)
\(\Rightarrow-\frac{2}{3}\cdot x=\frac{1}{10}\)
\(\Rightarrow x=\frac{1}{10}:\left(-\frac{2}{3}\right)\)
\(\Rightarrow x=-\frac{3}{20}\)
d) \(4-\left(\frac{1}{2}\cdot x+\frac{3}{4}\right)=-\frac{1}{5}\)
\(\Rightarrow\left(\frac{1}{2}\cdot x+\frac{3}{4}\right)=4-\left(-\frac{1}{5}\right)\)
\(\Rightarrow\)\(\frac{1}{2}\cdot x+\frac{3}{4}=\frac{21}{5}\)
\(\Rightarrow\)\(\frac{1}{2}\cdot x=\frac{21}{5}-\frac{3}{4}\)
\(\Rightarrow\)\(\frac{1}{2}\cdot x=\frac{69}{20}\)
\(\Rightarrow\)\(x=\frac{69}{20}:\frac{1}{2}\)
\(\Rightarrow\)\(x=\frac{69}{10}\)
a) x là số nguyên => x+1 là số nguyên
=> x+1 thuộc Ư (6)={-6;-3;-2;-1;1;2;3;6}
x+1 | -6 | -3 | -2 | -1 | 1 | 2 | 3 | 6 |
x | -7 | -4 | -3 | -2 | 0 | 1 | 2 | 5 |
b) y nguyên => y+3 nguyên
=> x; y+3 thuộc Ư (-8)={-8;-4;-2;-1;1;2;4;8}
x | -8 | -4 | -2 | -1 | 1 | 2 | 4 | 8 |
y+3 | 1 | 2 | 4 | 8 | -8 | -4 | -2 | -1 |
y | -2 | -1 | 1 | 5 | -11 | -7 | -5 | -4 |
c) xy-x+y=6
<=> x(y-1)+(y-1)=5
<=> (x+1)(y-1)=5
Vì x, y nguyên => x+1;y-1 nguyên => x+1; y-1 thuộc Ư (5)={-5;-1;1;5}
Ta có bảng
x+1 | -5 | -1 | 1 | 5 |
y-1 | -1 | -5 | 5 | 1 |
x | -6 | -2 | 0 | 4 |
y | 0 | -4 | 6 | 2 |
a,\(x+1\inƯ\left(6\right)\)
\(=>x+1\in\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
\(=>x\in\left\{-7;-4;-3;-2;0;1;2;5\right\}\)
b,\(\left(4x+3\right).\left(x-2\right)=0\)
\(=>\orbr{\begin{cases}4x+3=0\\x-2=0\end{cases}=>\orbr{\begin{cases}4x=-3\\x=2\end{cases}}}\)
\(=>\orbr{\begin{cases}x=\frac{-3}{4}\\x=2\end{cases}}\)
c,\(\left(15x+8\right).\left(12x-1\right)=0\)
\(=>\orbr{\begin{cases}15x+8=0\\12x-1=0\end{cases}}=>\orbr{\begin{cases}15x=-8\\12x=1\end{cases}}\)
\(=>\orbr{\begin{cases}x=\frac{-8}{15}\\x=\frac{1}{12}\end{cases}}\)
e,\(xy-x+y=6\)
\(=>x.\left(y-1\right)+y-1=5\)
\(=>\left(x+1\right).\left(y-1\right)=5\)
Ta có bảng sau :
tự lập bảng =))
a)x=-2,791287847 nhé