tim x: (x^2-1)(x^2-4)(x^2-7)(x^2-10)<0
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
x x - 4/7 = 15/20
(hình như sai đề, tại vì lớp 4 chưa học lũy thừa)
x * 2/7 -2/3 =1/2
x* 2/7 = 1/2 + 2/3
x* 2/7 = 7/6
x= 7/6 : 2/7
x= 49/12
x : 5/4 + 4/5 =2
x: 5/4 = 2 - 4/5
x: 5/4 = 6/5
x= 6/5 * 5/4
x= 3/2
4/3 x 7/8 x 5
= 7/6 x 5
= 35/6
8/9 x 6/5 x 2/3
= 16/15 x 2/3
= 32/45
4/7 : 5/7 x 10/11
= 4/5 x 10/11
= 8/11
8/13 : 7/13 : 4
= 8/7 : 4
= 2/7
\(x\cdot\frac{22}{9}=\frac{22}{5}-\frac{11}{7}\)
\(x\cdot\frac{22}{9}=\frac{99}{35}\)
\(x=\frac{99}{35}:\frac{22}{9}\)
\(x=\frac{81}{70}\)
Vậy \(x=\frac{81}{70}\).
\(x:\frac{56}{5}=\frac{26}{5}+\frac{52}{5}\)
\(x:\frac{56}{5}=\frac{78}{5}\)
\(x=\frac{78}{5}\cdot\frac{56}{5}\)
\(x=\frac{4368}{25}\)
Vậy \(x=\frac{4368}{25}\).
\(x:\frac{19}{3}=4\)
\(x=4\cdot\frac{19}{3}\)
\(x=\frac{76}{3}\)
Vậy \(x=\frac{76}{3}\).
b) (x+2)5=210 c) 5x : 52 = 125 d) (3x-24) . 73= 2x74 ( câu a khá dài nên mình để sau nhé )
(x+2)5=45 5x : 52 = 53 (3x-24) . 73= 2x74 : 73
x+2=4 5x = 53 . 52 3x-24 = 2x7
x=4-2 5x = 55 3x -16 =14
x=2 x = 5 3x=14+16
3x=30
x=30:3
x=10
a) (x-5)4 = (x-5)6
(x-5)4 - (x-5)6 =0
(x-5)4 - (x-5)4 . (x-5)2 =0
(x-5)4 - [ 1- (x-5 )2 ] =0
TH1: ( x-5 )4 =0 TH2: 1-( x-5)2 = 0
x-5=0 (x-5)2 = 1-0
x=0+5 (x-5)2 =1
x=5 x-5 = +-1
* x-5=1
x=6
* x-5=-1
x=4
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
= \(1-\frac{1}{7}\)
= \(\frac{7}{7}-\frac{1}{7}\)
= \(\frac{6}{7}\)
2) \(\frac{7}{4}-x.\frac{4}{3}=\frac{5}{19}\)
\(x.\frac{4}{3}=\frac{7}{4}-\frac{5}{19}\)
\(x.\frac{4}{3}=\frac{133}{76}-\frac{20}{76}\)
\(x.\frac{4}{3}=\frac{113}{76}\)
\(x=\frac{113}{76}:\frac{4}{3}\)
\(x=\frac{399}{304}\)
VẬY \(x=\frac{399}{304}\)
b) \(\left(x+\frac{3}{4}\right).\frac{5}{7}=\frac{10}{9}\)
\(\left(x+\frac{3}{4}\right)=\frac{10}{9}:\frac{5}{7}\)
\(x+\frac{3}{4}=\frac{14}{9}\)
\(x=\frac{14}{9}-\frac{3}{4}\)
\(x=\frac{29}{36}\)
Vậy \(x=\frac{29}{36}\)
c) \(x.\frac{1}{2}+\frac{3}{2}.x=\frac{4}{5}\)
\(x.\left(\frac{1}{2}+\frac{3}{2}\right)=\frac{4}{5}\)
\(x.2=\frac{4}{5}\)
\(x=\frac{4}{5}:2\)
\(x=\frac{2}{5}\)
Vậy \(x=\frac{2}{5}\)
Chúc bạn học tốt !!!
a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)
c: Ta có: \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b.
PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$
$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$
$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$
$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$
$\Leftrightarrow (x-3)^2(2x+1)^2=0$
$\Leftrightarrow (x-3)(2x+1)=0$
$\Leftrightarrow x-3=0$ hoặc $2x+1=0$
$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$
d.
$x^2-2x=24$
$\Leftrightarrow x^2-2x-24=0$
$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$
$\Leftrightarrow x=-4$ hoặc $x=6$