Câu 1: tìm x, biết:
a, 11 - ( 2x - 3 ) = 20142015 : 20142014 b, 1 + 1/3 + 1/6 + 1/10 +....+ 2/x(x+1) = 1.1993/1995
mn giups mk vs, trong đêm nay mk pải có bài, sáng mai mk hk rồi
Tìm x , biết:
a) |2x +1| - 3 = x +4
b) |3x - 5 | = 1 - 3x
c) |2x + 2|+|x - 1| = 10
d) |x - 3|+|x +4|+|2x + 6|=10
giúp mik vs ạ,nếu có gì sai xót trong đề thì mn thông cảm ạ
`|2x+1|-3=x+4`
`<=>|2x+1|=x+4+3=x+7(x>=-7)`
`**2x+1=x+7`
`<=>x=7-1=6(tm)`
`**2x+1=-x-7`
`<=>3x=-6`
`<=>x=-2(tm)`
`|3x-5|=1-3x(x<=1/3)`
`**3x-5=1-3x`
`<=>6x=6`
`<=>x=1(l)`
`**3x-5=3x-1`
`<=>-5=-1` vô lý
`|2x+2|+|x-1|=10`
Nếu `x>=1`
`pt<=>2x+2+x-1=10`
`<=>3x+1=10`
`<=>3x=9`
`<=>x=3(tm)`
Nếu `x<=-1`
`pt<=>-2x-2+1-x=10`
`<=>-1-3x=10`
`<=>-11=3x`
`<=>x=-11/3(tm)`
Nếu `-1<=x<=1`
`pt<=>2x+2+1-x=10`
`<=>x+3=10`
`<=>x=7(l)`
Vậy `S={3,-11/3}`
d)
+) Với \(x< -4\), PT \(\Rightarrow3-x-x-4-2x-6=10\) \(\Leftrightarrow x=-\dfrac{17}{4}\) (Nhận)
+) Với \(-4\le x\le-3\), PT \(\Rightarrow3-x+x+4-2x-6=10\) \(\Leftrightarrow x=-\dfrac{9}{2}\) (Loại)
+) Với \(-3< x\le3\), PT \(\Rightarrow3-x+x+4+2x+6=10\) \(\Leftrightarrow x=-\dfrac{3}{2}\) (Nhận)
+) Với \(x>3\), PT \(\Rightarrow x+3+x+4+2x+6=10\) \(\Leftrightarrow x=-\dfrac{3}{4}\) (Loại)
Vậy \(x\in\left\{-\dfrac{3}{2};-\dfrac{17}{4}\right\}\)
Tìm x, biết:
a) x3-1-(x2+2x)(x-2)=5
b) (x+1)3-(x-1)3-6(x-1)2=-10
a) x3-1-(x2+2x)(x-2)=5
⇔ x3-1-x3+4x=5
⇔ 4x=6
⇔ \(x=\dfrac{3}{2}\)
Giari ptr
a/2x(3x-1)=6x^2-13
b/\(\dfrac{x}{3}-\dfrac{2x+1}{6}=\dfrac{x}{6}-x\)
Giups mk vs ạ ai nhanh mk tick nha ><
a) \(6x^2-2x-6x^2+13=0\\ -2x=-13\\ x=\dfrac{13}{2}\)
b: =>2x-2x-1=x-6x
=>-5x=-1
hay x=1/5
Lời giải:
a.
$2x(3x-1)=6x^2-13$
$\Leftrightarrow 6x^2-2x=6x^2-13$
$\Leftrightarrow 2x=13$
$\Leftrightarrow x=\frac{13}{2}$
b.
$\frac{x}{3}-\frac{2x+1}{6}=\frac{x}{6}-x$
$\Leftrightarrow \frac{2x-(2x+1)}{6}=\frac{-5}{6}x$
$\Leftrightarrow \frac{-1}{6}=\frac{-5}{6}x$
$\Leftrightarrow x=\frac{-1}{6}: \frac{-5}{6}=\frac{1}{5}$
Tìm x biết:
a) x(x-3)+2x-6=0
b) (x+1)2-4(x+1)=0
c) (2x+5)(4x+3)-8x(x+3)=10
a: \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
b: \(\left(x+1\right)^2-4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
Bài 10: Tìm các số nguyên \(x\) biết:
a) \(2x-3\) là bội của \(x+1\)
b) \(x-2\) là ước của \(3x-2\)
Bài 14: Tìm số tự nhiên \(n\) sao cho:
a) \(4n-5\) ⋮ \(2n-1\)
b) \(n^2+3n+1\) ⋮ \(n+1\)
Bài 16: Tìm cặp số tự nhiên \(x\),\(y\) biết:
a) \(\left(x+5\right)\left(y-3\right)=15\)
b) \(\left(2x-1\right)\left(y+2\right)=24\)
c) \(xy+2x+3y=0\)
d) \(xy+x+y=30\)
Bài 10:
a: 2x-3 là bội của x+1
=>\(2x-3⋮x+1\)
=>\(2x+2-5⋮x+1\)
=>\(-5⋮x+1\)
=>\(x+1\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{0;-2;4;-6\right\}\)
b: x-2 là ước của 3x-2
=>\(3x-2⋮x-2\)
=>\(3x-6+4⋮x-2\)
=>\(4⋮x-2\)
=>\(x-2\inƯ\left(4\right)\)
=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(x\in\left\{3;1;4;0;6;-2\right\}\)
Bài 14:
a: \(4n-5⋮2n-1\)
=>\(4n-2-3⋮2n-1\)
=>\(-3⋮2n-1\)
=>\(2n-1\inƯ\left(-3\right)\)
=>\(2n-1\in\left\{1;-1;3;-3\right\}\)
=>\(2n\in\left\{2;0;4;-2\right\}\)
=>\(n\in\left\{1;0;2;-1\right\}\)
mà n>=0
nên \(n\in\left\{1;0;2\right\}\)
b: \(n^2+3n+1⋮n+1\)
=>\(n^2+n+2n+2-1⋮n+1\)
=>\(n\left(n+1\right)+2\left(n+1\right)-1⋮n+1\)
=>\(-1⋮n+1\)
=>\(n+1\in\left\{1;-1\right\}\)
=>\(n\in\left\{0;-2\right\}\)
mà n là số tự nhiên
nên n=0
Bài 16:
a: \(\left(x+5\right)\left(y-3\right)=15\)
=>\(\left(x+5\right)\left(y-3\right)=1\cdot15=15\cdot1=\left(-1\right)\cdot\left(-15\right)=\left(-15\right)\cdot\left(-1\right)=3\cdot5=5\cdot3=\left(-3\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-3\right)\)
=>\(\left(x+5;y-3\right)\in\left\{\left(1;15\right);\left(15;1\right);\left(-1;-15\right);\left(-15;-1\right);\left(3;5\right);\left(5;3\right);\left(-3;-5\right);\left(-5;-3\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(-4;18\right);\left(10;4\right);\left(-6;-12\right);\left(-20;2\right);\left(-2;8\right);\left(0;6\right);\left(-8;-2\right);\left(-10;0\right)\right\}\)
mà (x,y) là cặp số tự nhiên
nên \(\left(x,y\right)\in\left\{\left(10;4\right);\left(0;6\right)\right\}\)
b: x là số tự nhiên
=>2x-1 lẻ và 2x-1>=-1
\(\left(2x-1\right)\left(y+2\right)=24\)
mà 2x-1>=-1 và 2x-1 lẻ
nên \(\left(2x-1\right)\cdot\left(y+2\right)=\left(-1\right)\cdot\left(-24\right)=1\cdot24=3\cdot8\)
=>\(\left(2x-1;y+2\right)\in\left\{\left(-1;-24\right);\left(1;24\right);\left(3;8\right)\right\}\)
=>\(\left(2x;y\right)\in\left\{\left(0;-26\right);\left(2;22\right);\left(4;6\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(0;-26\right);\left(1;11\right);\left(2;6\right)\right\}\)
mà (x,y) là cặp số tự nhiên
nên \(\left(x,y\right)\in\left\{\left(1;11\right);\left(2;6\right)\right\}\)
c:
x,y là các số tự nhiên
=>x+3>=3 và y+2>=2
xy+2x+3y=0
=>\(xy+2x+3y+6=6\)
=>\(x\left(y+2\right)+3\left(y+2\right)=6\)
=>\(\left(x+3\right)\left(y+2\right)=6\)
mà x+3>=3 và y+2>=2
nên \(\left(x+3\right)\cdot\left(y+2\right)=3\cdot2\)
=>x=0 và y=0
d: xy+x+y=30
=>\(xy+x+y+1=31\)
=>\(x\left(y+1\right)+\left(y+1\right)=31\)
=>\(\left(x+1\right)\left(y+1\right)=31\)
\(\Leftrightarrow\left(x+1\right)\cdot\left(y+1\right)=1\cdot31=31\cdot1=\left(-1\right)\cdot\left(-31\right)=\left(-31\right)\cdot\left(-1\right)\)
=>\(\left(x+1;y+1\right)\in\left\{\left(1;31\right);\left(31;1\right);\left(-1;-31\right);\left(-31;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(0;30\right);\left(30;0\right);\left(-2;-32\right);\left(-32;-2\right)\right\}\)
mà (x,y) là cặp số tự nhiên
nên \(\left(x,y\right)\in\left\{\left(0;30\right);\left(30;0\right)\right\}\)
Tìm x biết:a) 10/x = -15/9; b) x/9 = -11/5 : 0,6; c) -7/8 - 2x = -3/4; d) (x-1/2):1/3+5/7=9và5/7; e)1/15.x+4/5.x=5và1/5; f) -1/3<x/6<1/2(x thuộc Z); h) 3/5+2/5:x=-1/4;i) 4 và 3/4x - 3 và 1/2=5/4; k)9/4.(1/3x-1/2)=4và1/2
Bài 4: Tìm x, biết:
a) 3(2x – 3) + 2(2 – x) = –3 ; b) x(5 – 2x) + 2x(x – 1) = 13 ;
c) 5x(x – 1) – (x + 2)(5x – 7) = 6 ; d) 3x(2x + 3) – (2x + 5)(3x – 2) = 8 ;
e) 2(5x – 8) – 3(4x – 5) = 4(3x – 4) + 11; f) 2x(6x – 2x 2 ) + 3x 2 (x – 4) = 8.
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy: \(x=\dfrac{1}{2}\)
===========
b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
\(\Leftrightarrow x=\dfrac{13}{3}\)
Vậy: \(x=\dfrac{13}{3}\)
==========
c/ \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)
\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)
\(\Leftrightarrow-2x=-2\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
\(\Leftrightarrow x=\dfrac{2}{7}\)
Vậy: \(x=\dfrac{2}{7}\)
==========
f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow-x^3=8\)
\(\Leftrightarrow x=-2\)
Vậy: \(x=-2\)
Tìm x biết:
a) \(\left|x+2\dfrac{1}{2}\right|=\left|3x+1\right|\)
b) \(\left|2x-6\right|+\left|x+3\right|=8\)
c) \(2.\left|x+2\right|+\left|4-x\right|=11\)
\(c,\Rightarrow\left[{}\begin{matrix}-2\left(x+2\right)+\left(4-x\right)=11\left(x< -2\right)\\2\left(x+2\right)+\left(4-x\right)=11\left(-2\le x\le4\right)\\2\left(x+2\right)+\left(x-4\right)=11\left(x>4\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{3}\left(tm\right)\\x=3\left(tm\right)\\x=\dfrac{11}{3}\left(ktm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{3}\end{matrix}\right.\)
\(a,\Rightarrow\left[{}\begin{matrix}x+\dfrac{5}{2}=3x+1\\x+\dfrac{5}{2}=-3x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{7}{8}\end{matrix}\right.\)
\(b,\Rightarrow\left[{}\begin{matrix}6-2x-x-3=8\left(x\le-3\right)\\6-2x+x+3=8\left(-3\le x\le3\right)\\2x-6+x+3=8\left(x>3\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-5}{3}\left(ktm\right)\\x=1\left(tm\right)\\x=\dfrac{11}{3}\left(tm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{11}{3}\end{matrix}\right.\)
Câu 2: Tìm x biết:
a. \(\sqrt{x-3}=5\)
b. \(\sqrt{2x-1}=\sqrt{3}\)
c. \(\sqrt{1-x}=-1\)
d. \(\sqrt{\left(x-1\right)^2}=1\)
\(a,ĐK:x\ge3\\ PT\Leftrightarrow x-3=5\Leftrightarrow x=8\left(tm\right)\\ b,ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow2x-1=3\Leftrightarrow x=2\left(tm\right)\\ c,Vì.\sqrt{1-x}\ge0>-1.nên.pt.vô.nghiệm\\ d,PT\Leftrightarrow\left|x-1\right|=1\Leftrightarrow\left[{}\begin{matrix}x-1=1\\1-x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
a) \(\sqrt{x-3}=5\) (1)
ĐKXĐ: \(x\ge3\)
\(\left(1\right)\Leftrightarrow x-3=25\)
\(\Leftrightarrow x=28\) (nhận)
Vậy \(x=28\)
b) \(\sqrt{2x-1}=\sqrt{3}\) (2)
ĐKXĐ: \(x\ge\dfrac{1}{2}\)
\(\left(2\right)\Leftrightarrow2x-1=3\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\) (nhận)
Vậy \(x=2\)
c) \(\sqrt{1-x}=-1\)
Không tìm được \(x\) vì \(\sqrt{1-x}\ge0\) (với mọi \(x\le1\))
d) \(\sqrt{\left(x-1\right)^2}=1\) (3)
ĐKXĐ: Với mọi \(x\in R\)
\(\left(3\right)\Leftrightarrow\left|x-1\right|=1\)
\(\Leftrightarrow x-1=1\) (khi \(x\ge1\)) hoặc \(1-x=1\) (khi \(x< 1\))
* \(x-1=1\)
\(\Leftrightarrow x=2\) (nhận)
* \(1-x=1\)
\(\Leftrightarrow x=0\) (nhận)
Vậy \(x=0;x=2\)