Bài 2: Tìm x
A) 2x+5/2=-3/5
B) 1/2:x-5/6=-2/3
C) (312-x):12,6=24,5
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a: x=5:(-1/2)=-10
b: x=8/3+1/9=25/9
c: =>x+5/6=11/21
=>x=-13/42
d: =>7/4x-5=-10/3
=>7/4x=5/3
=>x=20/21
e: =>10/3-3/4:x=-1/6
=>3/4:x=10/3+1/6=21/6=7/2
=>x=3/4:7/2=3/4*2/7=6/28=3/14
g: =>3/(x+5)=3/20
=>x+5=20
=>x=15
h: =>1-1/2+1/2-1/3+...+1/x-1/x+1=49/50
=>1-1/x+1=49/50
=>x+1=50
=>x=49
a) Ta có: \(\dfrac{1}{4}-\left|x+\dfrac{1}{2}\right|=\dfrac{1}{8}\)
\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|=\dfrac{1}{8}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{8}\\x+\dfrac{1}{2}=-\dfrac{1}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{8}\\x=\dfrac{-5}{8}\end{matrix}\right.\)
a) 13,104 : x = 6,88
x= 13,104 : 6,88
x= 819/430
b) ( 312 - x ) : 12,6 = 24,5
312 - x = 24,5 * 12,6
312 - x = 308,7
x = 312 - 308,7
x = 3,3
OK
a) x = 13,104 : 6,88
<=> x = 819 / 430
b) 312 - x = 24,5 x 12,6
312- x = 308,7
x = 308,7 - 312
x = - 3,3
a) \(\left(x-1\right)^3\)
\(=x^3-3x^2+3x-1\)
b) \(\left(2x-3y\right)^3\)
\(=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^3+\left(3y\right)^3\)
\(=8x^3-36x^2y+54xy^2-27y^3\)
Bài 3:
a: Ta có: \(\left(x-2\right)^3-x^2\left(x-6\right)=5\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=5\)
\(\Leftrightarrow12x=13\)
hay \(x=\dfrac{13}{12}\)
b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=4\)
\(\Leftrightarrow x^3-1-x^3+4x=4\)
\(\Leftrightarrow4x=5\)
hay \(x=\dfrac{5}{4}\)
Bài 1:
a: \(\Leftrightarrow x-1\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{2;0;4;-2\right\}\)
a: =2/5-3/5+3/7=3/7-1/5
=15/35-7/35
=8/35
b: =>5/7:x=4/3
=>x=5/7:4/3=5/7*3/4=15/28
c: =>x-1/3=15/8:4/5=15/8*5/4=75/32
=>x=75/32+1/3=257/96
d: =>2x+1/8=2/7
=>2x=9/56
=>x=9/112
e: =>2x=10/3-5/4-3/4=10/3-2=4/3
=>x=2/3
\(a,\dfrac{2}{5}+\dfrac{3}{7}+\left(-\dfrac{3}{5}\right)\\ =\dfrac{2}{5}+\dfrac{3}{7}-\dfrac{3}{5}\\=\left(\dfrac{2}{5}-\dfrac{3}{5}\right)+\dfrac{3}{7}\\ =-\dfrac{1}{5}+\dfrac{3}{7}\\ =-\dfrac{7}{35}+\dfrac{15}{35}\\ =\dfrac{8}{35}\\ b,1-\dfrac{5}{7}:x=-\dfrac{1}{3}\\ =>\dfrac{5}{7}:x=1-\left(-\dfrac{1}{3}\right)\\ =>\dfrac{5}{7}:x=1+\dfrac{1}{3}\\ =>\dfrac{5}{7}:x=\dfrac{3}{3}+\dfrac{1}{3}\\ =>\dfrac{5}{7}:x=\dfrac{4}{3}\\ =>x=\dfrac{5}{7}:\dfrac{4}{3}\\ =>x=\dfrac{5}{7}.\dfrac{3}{4}\\ =>x=\dfrac{15}{28}\\ c,\dfrac{4}{5}\left(x-\dfrac{1}{3}\right)=\dfrac{15}{8}\\ =>x-\dfrac{1}{3}=\dfrac{15}{8}:\dfrac{4}{5}\\ =>x-\dfrac{1}{3}=\dfrac{15}{8}.\dfrac{5}{4}\\ =>x-\dfrac{1}{3}=\dfrac{75}{32}\\ =>x=\dfrac{75}{32}+\dfrac{1}{3}\\ =>x=\dfrac{257}{96}\)
\(d,\dfrac{2}{3}:\left(2x+\dfrac{1}{8}\right)=\dfrac{7}{3}\\ =>2x+\dfrac{1}{8}=\dfrac{2}{3}:\dfrac{7}{3}\\ =>2x+\dfrac{1}{8}=\dfrac{2}{3}.\dfrac{3}{7}\\ =>2x+\dfrac{1}{8}=\dfrac{2}{7}\\ =>2x=\dfrac{2}{7}-\dfrac{1}{8}\\ =>2x=\dfrac{16}{56}-\dfrac{7}{56}\\ =>2x=\dfrac{9}{56}\\ =>x=\dfrac{9}{56}:2\\ =>x=\dfrac{9}{112}\\ e,2x+\dfrac{3}{4}=\dfrac{10}{3}-\dfrac{5}{4}\\ =>e,2x+\dfrac{3}{4}=\dfrac{40}{12}-\dfrac{15}{12}\\ =>2x+\dfrac{3}{4}=\dfrac{25}{12}\\ =>2x=\dfrac{25}{12}-\dfrac{3}{4}\\ =>2x=\dfrac{25}{12}-\dfrac{9}{12}\\ =>2x=\dfrac{16}{12}\\ =>2x=\dfrac{4}{3}\\ =>x=\dfrac{4}{3}:2\\ =>x=\dfrac{4}{6}\\ =>x=\dfrac{2}{3}\)
Lời giải:
a.
\(-16a^4b^6-24a^5b^5-9a^6b^4=-[(4a^2b^3)^2+2.(4a^2b^3).(3a^3b^2)+(3a^3b^2)^2]\)
\(=-(4a^2b^3+3a^3b^2)^2=-[a^2b^2(4b+3a)]^2\)
\(=-a^4b^4(3a+4b)^2\)
b.
$x^3-6x^2y+12xy^2-8x^3$
$=x^3-3.x^2.2y+3.x(2y)^2-(2y)^3=(x-2y)^3$
c.
$x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}$
$=x^3+3.x^2.\frac{1}{2}+3.x.\frac{1}{2^2}+(\frac{1}{2})^3$
$=(x+\frac{1}{2})^3$
a) Ta có: \(-16a^4b^6-24a^5b^5-9a^6b^4\)
\(=-a^4b^4\left(16b^2+24ab+9a^2\right)\)
\(=-a^4b^4\cdot\left(4b+3a\right)^2\)
b) Ta có: \(x^3-6x^2y+12xy^2-8y^3\)
\(=x^3-3\cdot x^2\cdot2y+3\cdot x\cdot\left(2y\right)^2-\left(2y\right)^3\)
\(=\left(x-2y\right)^3\)
c) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}\)
\(=x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3\)
\(=\left(x+\dfrac{1}{2}\right)^3\)
Bài 1:
A = 3(x + 1)2 + 5
Ta có: (x + 1)2 \(\ge\) 0 Với mọi x
\(\Rightarrow\) 3(x + 1)2 \(\ge\) 0 với mọi x
\(\Rightarrow\) 3(x + 1)2 + 5 \(\ge\) 5 với mọi x
Hay A \(\ge\) 5
Dấu "=" xảy ra khi và chỉ khi x + 1 = 5 hay x = -1
Vậy...
B = 2|x + y| + 3x2 - 10
Ta có: 2|x + y| \(\ge\) 0 với mọi x, y
3x2 \(\ge\) 0 với mọi x
\(\Rightarrow\) 2|x + y| + 3x2 - 10 \(\ge\) -10 với mọi x,y
Dấu "=" xảy ra khi và chỉ khi x + y = 0; x = 0
\(\Rightarrow\) x = y = 0
Vậy ...
C = 12(x - y)2 + x2 - 6
Ta có: 12(x - y)2 \(\ge\) 0 với mọi x; y
x2 \(\ge\) 0 với mọi x
\(\Rightarrow\) 12(x - y)2 + x2 - 6 \(\ge\) -6 với mọi x, y
Dấu "=" xảy ra khi và chỉ khi x = y = 0
Phần D ko rõ đầu bài nha vì D luôn có một giá trị duy nhất
Bài 2:
Phần A ko rõ đầu bài!
B = 3 - (x + 1)2 - 3(x + 2y)2
Ta có: -(x + 1)2 \(\le\) 0 với mọi x
-3(x + 2y)2 \(\le\) 0 với mọi x, y
\(\Rightarrow\) 3 - (x + 1)2 - 3(x + 2y)2 \(\le\) 3 với mọi x, y
Dấu "=" xảy ra khi và chỉ khi x = 2y; x + 1 = 0
\(\Rightarrow\) x = -1; y = \(\dfrac{-1}{2}\)
Vậy ...
C = -12 - 3|x + 1| - 2(y - 1)2
Ta có: -3|x + 1| \(\le\) 0 với mọi x
-2(y - 1)2 \(\le\) 0 với mọi y
\(\Rightarrow\) -12 - 3|x + 1| - 2(y - 1)2 \(\le\) -12 với mọi x, y
Dấu "=" xảy ra khi và chỉ khi x + 1 = 0; y - 1 = 0
\(\Rightarrow\) x = -1; y = 1
Vậy ...
Phần D đề ko rõ là \(\dfrac{5}{2x^2}-3\) hay \(\dfrac{5}{2}\)x2 - 3 nữa
F = \(\dfrac{-5}{3}\) - 2x2
Ta có: -2x2 \(\le\) 0 với mọi x
\(\Rightarrow\) \(\dfrac{-5}{3}-2x^2\) \(\le\) \(\dfrac{-5}{3}\) với mọi x
Dấu "=" xảy ra khi và chỉ khi x = 0
Vậy ...
Chúc bn học tốt!
\(a,2x+\dfrac{5}{2}=-\dfrac{3}{5}\)
\(2x=-\dfrac{3}{5}-\dfrac{5}{2}\)
\(2x=-\dfrac{31}{10}\)
\(x=-\dfrac{31}{10}:2\)
\(x=-\dfrac{31}{20}\)
\(b,\dfrac{1}{2}:x-\dfrac{5}{6}=-\dfrac{2}{3}\)
\(\dfrac{1}{2}:x=-\dfrac{2}{3}+\dfrac{5}{6}\)
\(\dfrac{1}{2}:x=\dfrac{1}{6}\)
\(x=\dfrac{1}{2}:\dfrac{1}{6}\)
\(x=3\)
\(c,\left(312-x\right):12,6=24,5\)
\(312-x=24,5\times12,6\)
\(312-x=308,7\)
\(x=312-308,7\)
`x=3,3`
a: =>2x=-3/5-5/2=-6/10-25/10=-31/10
=>x=-31/20
b: =>1/2:x=-2/3+5/6=5/6-4/6=1/6
=>x=1/2:1/6=3
c: =>312-x=308,7
=>x=3,3