Tìm x \(\in\)N sao cho:
\(a.6⋮\left(x-1\right)\)
\(b.14⋮\left(2x+3\right)\)
Cho các tập hợp sau A= \(\left\{x\in R|\left(x-2x^2\right)\left(x^2-3x+2\right)=0\right\}\) và B=\(\left\{n\in N|3< n\left(n+1\right)< 31\right\}\)
Tìm A \(\cap\) B
\(A=\left\{x\in R|\left(x-2x^2\right)\left(x^2-3x+2\right)=0\right\}\)
Giải phương trình sau :
\(\left(x-2x^2\right)\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow x\left(1-2x\right)\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\1-2x=0\\x-1=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\\x=2\end{matrix}\right.\)
\(\Rightarrow A=\left\{0;\dfrac{1}{2};1;2\right\}\)
\(B=\left\{n\in N|3< n\left(n+1\right)< 31\right\}\)
Giải bất phương trình sau :
\(3< n\left(n+1\right)< 31\)
\(\Leftrightarrow\left\{{}\begin{matrix}n\left(n+1\right)>3\\n\left(n+1\right)< 31\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}n^2+n-3>0\\n^2+n-31< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}n< \dfrac{-1-\sqrt[]{13}}{2}\cup n>\dfrac{-1+\sqrt[]{13}}{2}\\\dfrac{-1-5\sqrt[]{5}}{2}< n< \dfrac{-1+5\sqrt[]{5}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{-1-5\sqrt[]{5}}{2}< n< \dfrac{-1-\sqrt[]{13}}{2}\\\dfrac{-1+\sqrt[]{13}}{2}< n< \dfrac{-1+5\sqrt[]{5}}{2}\end{matrix}\right.\)
Vậy \(B=\left(\dfrac{-1-5\sqrt[]{5}}{2};\dfrac{-1-\sqrt[]{13}}{2}\right)\cup\left(\dfrac{-1+\sqrt[]{13}}{2};\dfrac{-1+5\sqrt[]{5}}{2}\right)\)
\(\Rightarrow A\cap B=\left\{2\right\}\)
2. tìm x
a) \(\left(x-1\right)^3=8\)
b) \(7^{2x-6}=49\)
c) \(\left(2x-14\right)^7=128\)
d) \(x^4.x^5=5^3.5^6\)
e) \(\left[3.\left(x+2\right):7\right].4=120\)
a) \(\left(x-1\right)^3=8=2^3\)
\(x-1=2\)
\(x=2+1=3\)
b) \(7^{2x-6}=49=7^2\)
\(2x-6=2\)
\(2x=6+2=8\)
\(x=8:2=4\)
c) \(\left(2x-14\right)^7=128=2^7\)
\(2x-14=2\)
\(2x=14+2=16\)
\(x=16:2=8\)
d) \(x^4\cdot x^5=5^3\cdot5^6=5^4\cdot5^5\)
\(x=5\)
e) \(3\cdot\left(x+2\right):7\cdot4=120\)
\(x+2=120:3\cdot7:4\)
\(x+2=70\)
\(x=70-2=68\)
Lời giải:
a. $(x-1)^3=8=2^3$
$\Rightarrow x-1=2$
$\Rightarrow x=3$
b. $7^{2x-6}=49=7^2$
$\Rightarrow 2x-6=2$
$\Rightarrow 2x=8$
$\Rightarrow x=4$
c. $(2x-14)^7=128=2^7$
$\Rightarrow 2x-14=2$
$\Rightarrow 2x=16$
$\Rightarrow x=18$
d.
$x^4.x^5=5^3.5^6$
$x^9=5^9$
$\Rightarrow x=5$
e.
$3(x+2):7=120:4=30$
$3(x+2)=30.7=210$
$x+2=210:3=70$
$x=70-2=68$
1/ Tìm các n \(\in\)Z thỏa: \(\left(n^2-1\right)\left(n^2-11\right)\left(n^2-21\right)\left(n^2-31\right)< 0\).
2/ Tìm các x \(\in\)Z sao cho: \(\left(4x-3\right)⋮\left(x-2\right)\).
3/ Tìm x, y \(\in\)Z biết: \(\left(2x-5\right)\left(y-6\right)=17\).
4/ Chứng minh: nếu a \(⋮\)b thì:
a/ \(a⋮\left(-b\right)\) b/ \(\left(a\right)⋮b\)và \(\left(-a\right)⋮\left(-b\right)\) c/ \(\left|a\right|⋮\left|b\right|\)
5/ Tìm các số nguyên n sao cho:
a/ \(n\left(n+4\right)< 0\) b/ \(\left(n+4\right)\left(5-n\right)< 0\)
6/ Chứng tỏ: \(\left(-1\right)a=-a\)
2/ Ta có : 4x - 3 \(⋮\) x - 2
<=> 4x - 8 + 5 \(⋮\) x - 2
<=> 4(x - 2) + 5 \(⋮\) x - 2
<=> 5 \(⋮\)x - 2
=> x - 2 thuộc Ư(5) = {-5;-1;1;5}
Ta có bảng :
x - 2 | -5 | -1 | 1 | 5 |
x | -3 | 1 | 3 | 7 |
Tìm các số nguyên x sao cho tích của 2 số hữu tỉ \(-\dfrac{3}{x-1};\dfrac{x-2}{2}\) là một số nguyên
Giải :
Ta có :
\(-\dfrac{3}{x-1}.\dfrac{x-2}{2}=\dfrac{-3\left(x-2\right)}{\left(x-1\right).2}=\dfrac{-3x+6}{2x-2}\)
\(\dfrac{-3x+6}{2x-2}\) là một số nguyên khi \(-3x+6⋮2x-2\)
\(\Leftrightarrow2\left(-3x+6\right)+3\left(2x-2\right)⋮2x-2\\ \Leftrightarrow-6x+12+6x-6⋮2x-2\\ \Leftrightarrow\left(-6x+6x\right)+\left(12-6\right)⋮2x-2\\ \Leftrightarrow6⋮2x-2\\ \Leftrightarrow2x-2\inƯ\left(6\right)=\left\{1;-1;2;-2;3;-3;6;-6\right\}\\ \Leftrightarrow2x\in\left\{3;1;4;0;5;-1;8;-4\right\}\\ \Leftrightarrow x\in\left\{2;0;4;-2\right\}\)
Cho \(E=\left\{x\in Z|\left|x\right|\le5\right\}\); \(A=\left\{x\in R|x^2+3x-4=0\right\}\);
\(B=\left\{x\in Z|(x-2)(x+1)(2x^2-x-3)=0\right\}\)
a) CM \(A\subset E\),\(B\subset E\)
b) Tìm \(E\backslash\left(A\cap B\right)\),\(E\backslash\left(A\cup B\right)\) rồi tìm quan hệ giữa hai tập hợp này.
\(E=\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)
\(A=\left\{1;-4\right\}\)
\(B=\left\{2;-1\right\}\)
a) Với mọi x thuộc A đều thuộc E \(\Rightarrow A\subset E\)
Với mọi x thuộc B đều thuộc E \(\Rightarrow B\subset E\)
b) \(A\cap B=\varnothing\)
\(\Rightarrow E\backslash\left(A\cap B\right)=\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)
\(A\cup B=\left\{-4;-1;1;2\right\}\)
\(\Rightarrow E\backslash\left(A\cup B\right)=\left\{-5;-3;-2;0;3;4;5\right\}\)
\(\Rightarrow E\backslash\left(A\cup B\right)\subset E\backslash\left(A\cap B\right)\)
Tìm x
a)\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)6
b)\(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
Nhanh nha kick cho :)
Ta có (x - 2)2 - (x - 3) (x + 3) = 6
=> (x - 2)2 - (x2 - 32) = 6
=> x2 - 4x + 22 - x2 + 32 = 6
=> x2 - x2 - 4x + 4 + 9 = 6
=> - 4x + 13 = 6
=> -4x = -7
=> x = \(\frac{-7}{-4}=\frac{7}{4}\)
a) \(\Leftrightarrow\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\Leftrightarrow x^2-4x+4-x^2+9=6\)
\(\Leftrightarrow13-4x=6\)
\(\Leftrightarrow4x=7\)
\(\Leftrightarrow x=\frac{7}{4}\)
b) \(4(x−3)^2−(2x−1)(2x+1)=10\)
\(\Leftrightarrow4\left(x^2-6x+9\right)−4x^2+1=10\)
\(\Leftrightarrow4x^2-24x+36−4x^2+1=10\)
\(\Leftrightarrow37-24x=10\)
\(\Leftrightarrow24x=27\)
\(\Leftrightarrow x=\frac{9}{8}\)
Tìm x, biết:
a, 2x-10-[3x-14-(4-5x)-2x] = 2
b,\(\left(\frac{1}{4}x-1\right)+\left(\frac{5}{6}x-2\right)-\left(\frac{3}{8}x+1\right)=4,5\)
a/ 2x - 10 - [3x - 14 - (4 - 5x) - 2x] = 2
=> 2x - 10 - (3x - 14 - 4 + 5x - 2x) = 2
=> 2x - 10 - 3x + 14 + 4 - 5x + 2x = 2
=> -4x + 6 = 0
=> -4x = -6
=> x = 3/2
b/ \(\left(\frac{1}{4}x-1\right)+\left(\frac{5}{6}x-2\right)-\left(\frac{3}{8}x+1\right)=4,5\)
\(\Rightarrow\frac{1}{4}x-1+\frac{5}{6}x-2-\frac{3}{8}x-1-\frac{9}{2}=0\)
\(\Rightarrow\frac{17}{24}x-\frac{17}{2}=0\)
\(\Rightarrow\frac{17}{24}x=\frac{17}{2}\)
\(\Rightarrow x=12\)
Bài 1: chứng minh rằng biểu thức sau không phụ thuộc vào x
a) A= (\(3x+7\))\(\left(2x-3\right)-\left(2x-5\right)^2-2x\left(x+6\right)+5-31x\)
b) B= \(2x\left(x-3\right)-2\left(x^2-3x\right)+14\)
c) C= \(\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
a: \(=6x^2-9x+14x-21-4x^2+20x-25-2x\left(x+6\right)+5-31x\)
\(=2x^2-6x-41-2x^2-12x\)
=-18x-41
b: \(=2x^2-6x-2x^2+6x+14=14\)
c: \(=x^3+1-x^3+1=2\)
Cho hai tập hợp \(A=\left\{\frac{3n}{n+1}n\in N,n< 4\right\}\)
\(B=\left\{x\in R,2x^3-x^2-6x=0\right\}\)
Tìm tất cả các tập X sao cho \(A\cap B\subset X\subset A\cup B\)