Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài làm
a) 64 . 4x = 168
=> 43 . 4x = ( 42 )8
=> 43 . 4x = 416
=> 43+x = 416
=> 3 + x = 16
=> x = 13
Vậy x = 13
b) x10 = 1x
Ta có: x10 = 1x
<=> x = 1
Hay 110 = 11
=> 1 = 1
Vậy x = 1
c) ( 7x - 11 )3 = 25 . 52 + 200
=> ( 7x - 11 )3 = 32 . 25 + 200
=> ( 7x - 11 )3 = 1000
=> ( 7x - 11 )3 = 103
=> 7x - 11 = 10
=> 7x = 21
=> x = 3
Vậy x = 3
# Học tốt #
Dạng 1:
a: =>x(x-3)=0
=>x=3 hoặc x=0
b: =>x(3x-4)=0
=>x=4/3 hoặc x=0
c: =>2x-1=0
=>x=1/2
d: =>2x(2x+3)=0
=>x=0 hoặc x=-3/2
e: =>x(2x+5)=0
=>x=-5/2 hoặc x=0
Lời giải:
a)
$M(x)=(x^5+5x^5)-2x^4-4x^3+3x$
$=6x^5-2x^4-4x^3+3x$
$N(x)=-6x^5+(7x^4-5x^4)+(x^3+3x^3)+4x^2-3x-1$
$=-6x^5+2x^4+4x^3+4x^2-3x-1$
b)
$M(-1)=6(-1)^5-2(-1)^4-4(-1)^3+3(-1)=-7$
$N(-2)=-6(-2)^5+2(-2)^4+4(-2)^3+4(-2)^2-3(-2)-1$
$=213$
c)
$M(x)+N(x)=(6x^5-2x^4-4x^3+3x)+(-6x^5+2x^4+4x^3+4x^2-3x-1)$
$=4x^2-1$
$M(x)-N(x)=(6x^5-2x^4-4x^3+3x)-(-6x^5+2x^4+4x^3+4x^2-3x-1)$
$=12x^5-4x^4-8x^3-4x^2+6x+1$
d)
$F(x)=M(x)+N(x)=4x^2-1=0\Leftrightarrow x^2=\frac{1}{4}$
$\Leftrightarrow x=\pm \frac{1}{2}$
Vậy $x=\pm \frac{1}{2}$ là nghiệm của $F(x)$
a. \(\frac{x}{2}=\frac{y}{3}=k\Rightarrow x=2k;y=3k\)
\(xy=54\Rightarrow2k3k=54\Rightarrow6k^2=54\Rightarrow k^2=9\Rightarrow k\in\left\{3;-3\right\}\)
\(k=3\Rightarrow x=6;y=9\)
\(k=-3\Rightarrow x=-6;y=-9\)
b.\(\frac{x}{5}=\frac{y}{3}=k\Rightarrow x=5k;y=3k\)
\(\Rightarrow\left(5k\right)^2-\left(3k\right)^2=4\Rightarrow25k^2-9k^2=4\)
\(\Rightarrow16k^2=4\Rightarrow k^2=\frac{1}{4}\Rightarrow k\in\left\{\frac{1}{2};-\frac{1}{2}\right\}\)
\(k=\frac{1}{2}\Rightarrow x=\frac{5}{2};y=\frac{3}{2}\)
\(k=-\frac{1}{2}\Rightarrow x=\frac{-5}{2};y=\frac{-3}{2}\)
c.\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{2}.\frac{1}{5}=\frac{y}{3}.\frac{1}{5}\Rightarrow\frac{x}{10}=\frac{y}{15}\)
\(\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{5}.\frac{1}{3}=\frac{z}{7}.\frac{1}{3}\Rightarrow\frac{y}{15}=\frac{z}{21}\)
\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x+y+z}{10+15+21}=\frac{92}{46}=2\)
\(\Rightarrow x=20,y=30,z=42\)
d.\(\frac{x^2}{9}=\frac{y^2}{16}\Rightarrow\frac{x^2}{9}=\frac{y^2}{16}=\frac{x^2+y^2}{9+16}=\frac{100}{25}=4\)
\(\Rightarrow x^2=36\Rightarrow x\in\left\{6;-6\right\};y^2=64\Rightarrow y\in\left\{8;-8\right\}\)
\(M\left(x\right)=P\left(x\right)+Q\left(x\right)=2,5x^6-4+2,5x^5-6x^3+2x^2\)-5x+\(3x-2,5x^6-x^2+5-2,5x^5+6x^3\)
=\(\left(2,5x^6-2,5x^6\right)\)+\(\left(2,5x^5-2,5x^5\right)\)\(\left(-6x^3+6x^3\right)\)+\(\left(2x^2-x^2\right)\)+\(\left(-5x+3x\right)\)+(-4+5)
= \(x^2-2x+1\)
a) \(2x^2-3x=0\)
\(\Leftrightarrow x\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
b) \(x^3-2x=0\)
\(\Leftrightarrow x\left(x^2-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{2}\end{matrix}\right.\)
c) \(x^6+1=0\)
\(\Leftrightarrow x^6=-1\)
Ta có : \(x^6\ge0\) với mọi x
Mà : -1 < 0
=> Vô nghiệm
d) \(x^3+2x=0\)
\(\Leftrightarrow x\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=-2\left(loại\right)\end{matrix}\right.\)
e) \(x^5+8x^2=0\)
\(\Leftrightarrow x^2\left(x^3+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x^3+8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^3=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
f) \(x^2\left(x^2-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x^2-9=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm3\end{matrix}\right.\)
g) \(\left(x+\dfrac{1}{2}\right)\left(x^2-\dfrac{4}{5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\x^2-\dfrac{4}{5}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x^2=\dfrac{4}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\sqrt{\dfrac{4}{5}}\end{matrix}\right.\)
\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)
\(=\left(\frac{x}{3}\right)^2=2\left(\frac{y}{4}\right)^2=4\left(\frac{z}{5}\right)^2\)
\(=\frac{x^2}{9}=\frac{2y^2}{16}=\frac{4z^2}{25}=\frac{x^2+2y^2+4z^2}{9+16+25}=\frac{141}{50}=2,82\)
Bạn tự => x , y , z nha
Ta có:
a) \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\Leftrightarrow\frac{x^2}{9}=\frac{2y^2}{32}=\frac{4z^2}{100}\) và \(x^2+2y^2+4z^2=141\)
Áp dụng t/chất dãy tỉ số bằng nhau ta được:
\(\frac{x^2}{9}=\frac{2y^2}{32}=\frac{4z^2}{100}=\frac{x^2+2y^2+4z^2}{9+32+100}=\frac{141}{141}=1\)
\(\Rightarrow x^2=9\); \(2y^2=32\) ; \(4z^2=100\)
\(\Rightarrow\hept{\begin{cases}x=-3\\y=-4\\z=-5\end{cases}}\) hoặc \(\hept{\begin{cases}x=3\\y=4\\z=5\end{cases}}\)
c) \(\frac{x}{10}=\frac{y}{6}=\frac{z}{24}\Leftrightarrow\frac{5x}{50}=\frac{y}{6}=\frac{-2z}{-48}\) và \(5x+y-2x=28\)
Áp dụng t/chất dãy tỉ số bằng nhau ta được:
\(\frac{5x}{50}=\frac{y}{6}=\frac{-2z}{-48}=\frac{5x+y-2z}{50+6-48}=\frac{28}{8}=\frac{7}{2}\)
\(\Rightarrow\hept{\begin{cases}5x=175\\y=21\\-2z=-168\end{cases}\Rightarrow\hept{\begin{cases}x=35\\y=21\\z=84\end{cases}}}\)
d) \(\frac{x}{3}=\frac{y}{4}\Leftrightarrow\frac{x}{15}=\frac{y}{20}\) và \(\frac{y}{5}=\frac{z}{7}\Leftrightarrow\frac{y}{20}=\frac{z}{28}\)
\(\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\) và theo đề ta có: 2x+3y-z = 186
\(\Leftrightarrow\hept{\begin{cases}\frac{2x}{30}=\frac{3y}{60}=\frac{-z}{-28}\\2x+3y-z=186\end{cases}}\)
Áp dụng t/chất dãy tỉ số bằng nhau ta được:
\(\frac{2x}{30}=\frac{3y}{60}=\frac{-z}{-28}=\frac{2x+3y-z}{30+60-28}=\frac{186}{62}=3\)
\(\Rightarrow\hept{\begin{cases}2x=90\\3y=180\\-z=-84\end{cases}\Rightarrow\hept{\begin{cases}x=45\\y=60\\z=84\end{cases}}}\)
b)\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\Leftrightarrow\frac{-2x^2}{-18}=\frac{y^2}{16}=\frac{-3z^2}{-75}\) và \(-2x^2+y^2-3z^2=-77\)
Áp dụng t/chất dãy tỉ số bằng nhau ta được:
\(\frac{-2x^2}{-18}=\frac{y^2}{16}=\frac{-3z^2}{-75}=\frac{-2x^2+y^2-3z^2}{-18+16-75}=\frac{-77}{-77}=1\)
\(\Rightarrow\hept{\begin{cases}-2x^2=-18\\y^2=16\\-3z^2=-75\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=4\\z=5\end{cases}}}\) hoặc \(\hept{\begin{cases}x=-3\\y=-4\\z=-5\end{cases}}\)
nha!!
a/ \(x+\dfrac{3}{5}=\dfrac{4}{7}\)
\(x=\dfrac{4}{7}-\dfrac{3}{5}\)
\(x=-\dfrac{1}{35}\)
Vậy ....
b/ \(x-\dfrac{5}{6}=\dfrac{1}{6}\)
\(x=\dfrac{1}{6}+\dfrac{5}{6}\)
\(x=1\)
Vậy ....
c/\(-\dfrac{5}{7}-x=\dfrac{-9}{10}\)
\(x=\dfrac{-5}{7}-\dfrac{-9}{10}\)
\(x=\dfrac{13}{70}\)
Vậy .....
d/ \(\dfrac{5}{7}-x=10\)
\(x=\dfrac{5}{7}-10\)
\(x=\dfrac{-65}{7}\)
Vậy ...
e/ \(x:\left(\dfrac{1}{9}-\dfrac{2}{5}\right)=\dfrac{-1}{2}\)
\(x:\dfrac{-13}{45}=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}.\dfrac{-13}{45}\)
\(x=\dfrac{13}{90}\)
Vậy ....
f/ \(\left(\dfrac{-3}{5}+1,25\right)x=\dfrac{1}{3}\)
\(0,65.x=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}:0,65\)
\(x=\dfrac{20}{39}\)
Vậy ....
g/ \(\dfrac{1}{3}x+\left(\dfrac{2}{3}-\dfrac{4}{9}\right)=\dfrac{-3}{4}\)
\(\dfrac{1}{3}x+\dfrac{2}{9}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{3}x=\dfrac{-35}{36}\)
\(\Leftrightarrow x=\dfrac{-35}{12}\)
Vậy ...
a) \(\left|x\right|=2,1\)
x= +- 2,1
b) \(\left|x\right|=\frac{3}{4}\left(x< 0\right)\)
x= -3/4
c) \(\left|x\right|=-1\frac{2}{5}\)
\(x\in\varphi\)
d) \(\left|x\right|=0,35\left(x>0\right)\)
\(x=0,35\)
a) |x| = 2,1 <=> \(\orbr{\begin{cases}x=2,1\\x=-2,1\end{cases}}\)
b) |x| = 3/4 <=> x = - 3/4 ( do x < 0 )
c) ko tim dc x vi |x| >= 0 voi moi x
d) |x| = 0,35 <=> x = 0,35 ( do x>0 )
Bài 1:
\(A=\left(x^3.x^3.x^2\right).\left(y.y^4\right).\left(\frac{2}{5}.\frac{-5}{4}\right)\)
\(A=x^8.y^5.\left(-\frac{1}{2}\right)\)
\(B=\left(x^5.x.x^2\right).\left(y^4.y^2.y\right).\left(\frac{-3}{4}.\frac{-8}{9}\right)\)
\(B=x^8.y^7.\frac{2}{3}\)
Bài 2:
\(A=\left(15.x^2.y^3-12.x^2.y^3\right)+\left(11x^3.y^2-8.x^3.y^2\right)+\left(7x^2-12x^2\right)\)
\(A=3.x^2.y^3+2.x^3.y^2-5x^2\)
B tương tự nhé, đáp án là (theo mình)
\(B=\frac{5}{2}.x^5.y+\frac{7}{3}.x.y^4-\frac{1}{4}.x^2.y^3\)
a: M(x)=x^2+2x-5+x^2-9x+5=2x^2-7x
N(x)=P(x)-Q(x)
=x^2+2x-5-x^2+9x-5=11x-10
b: M(x)=0
=>x(2x-7)=0
=>x=0 hoặc x=7/2
N(x)=0
=>11x-10=0
=>x=10/11
\(\text{#TNam}\)
`a,`
`M(x)=P(x)+Q(x)`
`M(x)= (x^2 + 2x − 5)+(x^2 − 9x + 5)`
`M(x)=x^2 + 2x − 5+x^2 − 9x + 5`
`M(x)= (x^2+x^2)+(2x-9x)+(-5+5)`
`M(x)=2x^2-7x`
`N(x)=(x^2 + 2x − 5)-(x^2 − 9x + 5)`
`N(x)=x^2 + 2x − 5-x^2 + 9x - 5`
`N(x)=(x^2-x^2)+(2x+9x)+(-5-5)`
`N(x)=11x-10`
`b,`
Đặt `M(x)=2x^2-7x=0`
`2x*x-7x=0`
`-> x(2x-7)=0`
`->`\(\left[{}\begin{matrix}x=0\\2x-7=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\2x=7\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x=\dfrac{7}{2}\end{matrix}\right.\)
Vậy, nghiệm của đa thức là `x=0, x=7/2`
Đặt `N(x)=11x-10=0`
`11x=0+10`
`11x=10`
`-> x=10 \div 11`
`-> x=10/11`
Vậy, nghiệm của đa thức là `x=10/11`