Tìm x thuộc số nguyên để:
a)\(\frac{1-x}{x+4}\inℤ\)
b) \(\frac{11-2x}{x-5}\inℤ\)
c) \(\frac{x+1}{2x+1}\inℤ\)
giúp mik với,tks
tìm các số nguyên x (x \(\inℤ\)) để:
a)M=\(\frac{x+3}{2}\)\(\inℤ\)
b)N=\(\frac{7}{x-1}\inℤ\)
c)P=\(\frac{x-1}{x+1}\inℤ\)
Giúp mik nhé.Thanks các bạn.
a) Ta có:
Để M = \(\frac{x+3}{2}\)\(\in\)Z <=> \(x+3⋮2\) <=> \(x+3\in\)B(2) = {0; 2; 4; ....}
<=> \(x\in\){-3; -1; 1; ....}
b) Để N = \(\frac{7}{x-1}\)\(\in\)Z <=> \(7⋮x-1\) <=> \(x-1\in\)Ư(7) = {1; -1; 7; -7}
Lập bảng :
x - 1 | 1 | -1 | 7 | -7 |
x | 2 | 0 | 8 | -6 |
Vậy ...
c) Ta có: P = \(\frac{x-1}{x+1}=\frac{x+1-2}{x+1}=1-\frac{2}{x+1}\)
Để P \(\in\)Z <=> \(2⋮x+1\) <=> \(x+1\in\)Ư(2) = {1; -1; 2; -2}
Lập bảng:
x + 1 | 1 | -1 | 2 | -2 |
x | 0 | -2 | 1 | -3 |
Vậy ...
để M nguyên thì \(\frac{x+3}{2}\) nguyên
=> (x+3) \(\in\)Ư(2)={-2:-1:1:2}
lập bảng ra tìm x nha bn ~!!
mấy ý kia tương tự !
a) \(M=\frac{x+3}{2}\in Z\)
=> x+3 chia hết cho 2
=> x+3 thuộc Ư(2)={-1,-2,1,2}
=> x thuộc {-4,-5,-2,-1}
b) \(N=\frac{7}{x-1}\in Z\)
=> x-1 thuộc Ư(7)={-1,-7,1,7}
=> x thuộc {0,-6,2,8}
Bài 1
\(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{x.\left(x+1\right)}=\frac{49}{50}\)
\(\frac{2x+3}{x-1}\)có giá trị là số nguyên \(\left(x\inℤ,x\ne0\right)\)
\(\frac{x-4}{y-3}=\frac{4}{3}\)và \(x-y=5\)\(\left(y\ne3\right)\)
Tìm x,y nguyên dương để: \(\frac{1}{x}+\frac{y}{2}=\frac{5}{8}\)
\(\left(x+3\right)^2+\left(y-1\right)^2< 4\left(x;y\inℤ\right)\)
\(\left(x+3\right)^2.\left(y-3\right)=-4\left(x;y\inℤ\right)\)
đổi k ko,mk hứa sẽ k lại(nếu ko làm chó!!!!!!!!!!!!!)
Bài 1: <Cho là câu a đi>:
a. \(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{49}{50}\)
\(\rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{49}{50}\)
\(\rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{49}{50}\)
\(\rightarrow1-\frac{1}{x+1}=\frac{49}{50}\)
\(\rightarrow\frac{1}{x+1}=1-\frac{49}{50}=\frac{1}{50}\)
\(\rightarrow x+1=50\rightarrow x=49\)
Vậy x = 49.
Tìm x, y nguyên dương để : \(\frac{1}{x}+\frac{y}{2}=\frac{5}{8}\)
Ta có : \(\frac{1}{x}+\frac{y}{2}=\frac{5}{8}\) => \(\frac{5}{8}-\frac{y}{2}=\frac{1}{x}\)
=> \(\frac{5-4y}{8}=\frac{1}{x}\) => \(\left(5-4y\right)x=8\)
=> 5 - 4y; x là ước của 8
Ta có bảng :
5 - 4y | 1 | 2 | 4 | 8 |
x | 8 | 4 | 2 | 1 |
y | 1 | 3/4 | 1/4 | -3/4 |
Vì x,y nguyên dương => x = 8 ; y = 1
Vậy x = 8; y = 1 là 2 giá trị cần tìm
Study well ! >_<
\(B=\dfrac{x-5}{x+1}\inℤ\)
\(C=\dfrac{2x+3}{x-2}\inℤ\)
\(D=\dfrac{6x-12}{2x+1}\inℤ\)
Tìm x,y,\(\inℤ\)
A
\(\frac{X-2}{4}=\frac{-9}{2-x}\)
B
\(\frac{3}{x+2}=\frac{5}{2x+1}\)
C\(\frac{3}{x+2}=\frac{5}{2x+1}\)
\(\frac{x-2}{4}=\frac{-9}{2-x}\)
\(\Rightarrow\frac{x-2}{4}=\frac{9}{x-2}\)
\(\Rightarrow\left(x-2\right)^2=36\)
\(\Rightarrow\left(x-2\right)^2=\left(\pm6\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=6\\x-2=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-4\end{cases}}}\)
\(\frac{3}{x+2}=\frac{5}{2x+1}\)
\(\Rightarrow3\left(2x+1\right)=\left(x+2\right)5\)
\(\Rightarrow6x+3=5x+10\)
\(\Rightarrow6x-5x=10-3\)
\(\Rightarrow x=7\)
c;giống câu trên :v
Tìm \(x\inℤ\),biết :
\(a)|x+1|=3\)
\(b)3^2x+2^4=5^2\)
\(c)\frac{4+x}{7+x}=\frac{4}{7}\)và \(x+y=11\)
a) \(|x+1|=3\)
\(\Rightarrow x+1=\pm3\)
+) \(x+1=3\) +) \(x+1=-3\)
\(\Rightarrow x=2\) \(\Rightarrow x=-4\)
Vậy \(x\in\left\{2;-4\right\}\)
b) \(3^2x+2^4=5^2\)
\(9x+16=25\)
\(9x=25-16\)
\(9x=9\)
\(x=1\)
c) \(\frac{4+x}{7+y}=\frac{4}{7}\)
\(\Rightarrow\left(4+x\right).7=\left(7+y\right).4\)
\(\Rightarrow28+7x=28+4y\)
\(\Rightarrow7x=4y\)
Mà \(\left(7,4\right)=1\) và \(x+y=11\)
Vậy \(x=4;y=7\)
a) Ta có: \(\left|x+1\right|=3\)
\(\Rightarrow x+1=\pm3\)
Nếu x + 1 = 3 => x = 2
Nếu x + 1 = -3 => x = -4
Vậy x = {2;-4}
b) \(3^2x+2^4=5^2\)
\(\Rightarrow9x+16=25\)
\(\Rightarrow9x=9\)
\(\Rightarrow x=1\)
Vậy x = 1
c) \(\frac{4+x}{7+x}=\frac{4}{7}\)
\(\Rightarrow7\left(4+x\right)=4\left(7+x\right)\)
\(\Rightarrow28+7x=28+4x\)
\(\Rightarrow7x-4x=0\)
\(\Rightarrow x=0\)
Vậy x = 0
a, Th1 : với \(x\ge-1\)
\(x+1=3\)=> \(x=2\)(t/m)
Th2: với \(x< -1\)
\(-x-1=3\)=> \(x=-4\)(t/m)
Tìm \(x\inℤ\)để\(A\inℤ\)
A=\(\frac{x+1}{x-2}\)( \(x\ne2\))
B=\(\frac{x-1}{x-2}\)(\(x\ne2\))
C=\(\frac{x+1}{x+2}\)(\(x\ne-2\))
D=\(\frac{2x-1}{x+1}\)(\(x\ne-1\))
Cho biểu thức: \(P=\frac{5x+1}{2x-3}.\frac{x+2}{25x^2-1}-\frac{8-3x}{25x^2-1}.\frac{5x+1}{2x-3}\)
a) Rút gọn P
b) Tìm x \(\inℤ\)để P có giá trị nguyên
a, \(ĐKXĐ:x\ne\pm\frac{1}{5},x\ne\frac{3}{2}\)
\(\Rightarrow P=\frac{\left(5x+1\right)\left(x+2\right)}{\left(2x-3\right)\left(5x-1\right)\left(5x+1\right)}-\frac{\left(8-3x\right)\left(5x+1\right)}{\left(5x-1\right)\left(5x+1\right)\left(2x-3\right)}\)
\(=\frac{x+2}{\left(2x-3\right)\left(5x-1\right)}-\frac{8-3x}{\left(5x-1\right)\left(2x-3\right)}\)
\(=\frac{2\left(2x-3\right)}{\left(2x-3\right)\left(5x-1\right)}=\frac{2}{5x-1}\)
b, Để P có giá trị nguyên thì \(2⋮5x-1\)
\(\Rightarrow5x-1\in\left\{1,2,-1,-2\right\}\)
=> x=..............
ĐKXĐ : x \(\ne\frac{3}{2}\) ; \(x\ne\frac{1}{5};x\ne-\frac{1}{5}\)
P= \(\frac{5x+1}{2x-3}.\left(\frac{x+2}{25x^2-1}-\frac{8-3x}{25x^2-1}\right)\)
P= \(\frac{5x-1}{2x-3}.\left(\frac{4x-6}{\left(5x+1\right).\left(5x-1\right)}\right)\)
P= \(\frac{5x-1}{2x-3}.\frac{2\left(2x-3\right)}{\left(5x-1\right)\left(5x+1\right)}\)
P= \(\frac{2}{5x-1}\)
KL
b,
Ta có \(x\inℤ\Rightarrow\frac{2}{5x-1}\inℤ\) \(\Leftrightarrow5x-1\inƯ(5)\) mà Ư(5)={+-1;+-5}
Tìm \(x\inℤ\), biết :
\(\frac {5}2x+\frac 1{2}x=x+400\)\(\%\).
\(\frac{5}{2}x+\frac{1}{2}x=x+400\%\)
\(\Rightarrow\left(\frac{5}{2}+\frac{1}{2}\right)x=x+4\)
\(\Rightarrow\frac{6}{2}x=x+4\)
\(\Rightarrow3x=x+4\)
\(\Rightarrow3x-x=4\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
Chúc bạn học tốt !!!