Tìm x,biết:
x(x+1)-(x-1)(x+2)=8
tìm các số thực x, y, z biết:
x + y + z + 8 = \(2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
\(x+y+z+8=2\sqrt[]{x-1}+4\sqrt[]{y-2}+6\sqrt[]{z-3}\left(1\right)\)
Áp dụng Bđt Bunhiacopxki :
\(\left(2\sqrt[]{x-1}+4\sqrt[]{y-2}+6\sqrt[]{z-3}\right)^2\le\left(2^2+4^2+6^2\right)\left(x-1+y-2+z-3\right)\)
\(\Leftrightarrow\left(2\sqrt[]{x-1}+4\sqrt[]{y-2}+6\sqrt[]{z-3}\right)^2\le56^{ }\left(x+y+z-6\right)\)
\(\Leftrightarrow\left(2\sqrt[]{x-1}+4\sqrt[]{y-2}+6\sqrt[]{z-3}\right)^2\le56^{ }\left(x+y+z+8\right)-784\)
Dấu "=" xảy ra khi và chỉ khi
\(\dfrac{x-1}{2}=\dfrac{y-2}{4}=\dfrac{z-3}{8}=\dfrac{x+y+z-6}{14}\left(2\right)\)
Đặt \(t=x+y+z+8\)
\(\left(1\right)\Leftrightarrow t^2=56t-784\)
\(\Leftrightarrow t^2-56t+784=0\)
\(\Leftrightarrow\left(t-28\right)^2=0\)
\(\Leftrightarrow t=28\)
\(\Leftrightarrow x+y+z+8=28\)
\(\Leftrightarrow x+y+z-6=14\)
\(\left(2\right)\Leftrightarrow\dfrac{x-1}{2}=\dfrac{y-2}{4}=\dfrac{z-3}{8}=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1.2=2\\y-2=1.4=4\\z-2=1.8=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=6\\z=10\end{matrix}\right.\) thỏa mãn đề bài
Tìm x, biết:
x+(x+1)+(x+2)+...+(x+30)=1240
`x+(x+1)+(x+2)+...+(x+30)=1240`
`=> (x + x + x + ... + x) + (1 + 2 + 3 +... + 30) = 1240`
`=> 31x + 465 = 1240`
`=> 31 x = 1240 - 465`
`⇒ 31x = 775`
`⇒ x = 775 : 31`
`⇒ x = 25`
\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\\ \left(x+x+...+x\right)+\left(1+2+...+30\right)=1240\\ 31x+465=1240\\ 31x=1240-465\\ 31x=775\\ x=775:31\\ x=25\)
\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\)
\(\Leftrightarrow\left(x+x+...+x\right)+\left(1+2+...+30\right)=1240\)
\(\Leftrightarrow31x+465=1240\)
\(\Leftrightarrow x=25\)
tìm x biết:x-1/x+2=x-2/x+3
x³ - x² - x = 1/3
<=> x³ = x² + x + 1/3
<=> 3x³ = 3(x² + x + 1/3)
<=> 3x³ = 3x² + 3x + 1
<=> 3x³ + x³ = x³ + 3x² + 3x + 1
<=> 4x³ = (x + 1)³
<=> ³√(4x³) = ³√(x + 1)³
<=> ³√4.x = x + 1
<=> ³√4.x - x = 1
<=> x(³√4 - 1) = 1
<=> x = 1/(³√4 - 1)
Ta có \(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)
\(\Rightarrow\left(x-1\right)\left(x+3\right)=\left(x+2\right)\left(x-2\right)\)
\(\Rightarrow x^2+2x-3=x^2-4\)
\(\Rightarrow x^2-x^2+2x=-4+3\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\frac{1}{2}\)
Vậy \(x=-\frac{1}{2}\)
Tìm X, biết:
X + (X + 1) + (X + 2) + (X + 3) + ...+ (X + 19) = 950
=>20x+190=950
=>20x=760
hay x=38
`20x+190=950`
`20x=760`
`x= 760: 20`
`x= 38`
20 x X = 950 - (1 + 2 + 3 + 4 + 5 + ... + 19)
20 x X = 950 - 190
20 x X = 760
X = 760 : 20
X = 38
tìm x biết:x+1/2+x+1/3+x+1/4+x+1/5=x+1/6
\(\Rightarrow x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}+x+\frac{1}{5}-x+\frac{1}{6}=0\)
\(\Rightarrow3x+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)
k cho minh
\(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}+x+\frac{1}{5}=x+\frac{1}{6}\)
\(\Leftrightarrow x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}+x+\frac{1}{5}-x-\frac{1}{6}=0\)
\(\Leftrightarrow3x+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}=0\)
Tính ra nhé !
\(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}+x+\frac{1}{5}=x+\frac{1}{6}\)
\(\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)=x+\frac{1}{6}\)
\(\Rightarrow4x+\frac{77}{60}=x+\frac{1}{6}\)
\(\Rightarrow3x=\frac{1}{6}-\frac{77}{60}\)
\(\Rightarrow3x=-\frac{67}{60}\)
\(\Rightarrow x=-\frac{67}{60}\div3=\frac{-67}{60.3}=-\frac{67}{180}\)
Vậy x = .........
tìm X biết:X+1+2+3+4+5-6-7-8-9=1-2-3-4-5+6+7+8+9
X + 1+2+3+4+5-6-7-8-9=1-2-3-4-5+6+7+8+9
X+ (-15) = 17
X = 17-(-15)
X = 32
vậy x = 32
tk nha
Tìm x biết:x+(x-1)+(x-2)+.......+(x-50)=255
x+(x-1)+(x-2)+....+(x-50)=225
<=>x+x-1+x-2+...+x-50=225
<=>51x-1275=225
<=>51x=-1050
bạn nè, câu trả lời của Nguyễn Quốc Khánh là sai đó
Tìm x ϵ Z biết:
x+(x+1)+(x+2)+...+2001+2002=2002
Lời giải:
Dãy $x,x+1, x+2,..., 2002$ có số số hạng là:
$\frac{2002-x}{1}+1=2003-x$
Tổng $x+(x+1)+....+2001+2002=\frac{(2002+x)(2003-x)}{2}$
Do đó:
$\frac{(2002+x)(2003-x)}{2}=2002$
$\Rightarrow (2002+x)(2003-x)=4004$
$2002.2003+x-x^2=4004$
$x^2-x-4006002=0$
$(x-2002)(x+2001)=0$
$\Rightarrow x=2002$ hoặc $x=-2001$
tìm x;y biết:x^3+x^2+x+1=y^3
\(x^2\left(x+1\right)+\left(x+1\right)=y^3\)
\(\left(x+1\right)\left(x^2+1\right)=y^3\)
\(\left(x+1\right)\left(x^2+1\right)-y^3=0\)
\(\orbr{\begin{cases}x+1=0\\x^2+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x^2=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\kothoaman\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}x=-1\\y^3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\y=0\end{cases}}\)
Vậy x = -1, y =0