Tìm x:
|2x-3|=x+1?
Help me!!!!!!!!
Tìm x biết : |2x+1|+|x-3|=5
Help me! Help me!!!!
help me
(x-3).(x^4+2x^2+1)=0. Tìm x?
vì \(x^4+2x^2+1=\left(x^2+1\right)^2\) mà \(x^2\ge0\Rightarrow x^2+1>0\Rightarrow\left(x^2+1\right)^2>0\)với mọi x.Nên x-3=0 .Từ đó suy ra x=3
Tìm x:
\(\dfrac{\sqrt{x}+3}{\sqrt{x}-1}\left(\dfrac{-2x+6}{\sqrt{x}-1}\right)=0\)
Help me plsss
ĐKXĐ: x>=0; x<>1
PT =>\(\dfrac{\left(\sqrt{x}+3\right)\left(-2x+6\right)}{\left(\sqrt{x}-1\right)^2}=0\)
=>6-2x=0
=>x=3
Tìm GTNN của hàm số y=\(\sqrt[3]{x^4+2x^2+1}\) - \(\sqrt[3]{x^2+1}+1\)
help me
Đặt \(\sqrt[3]{x^2+1}=t\left(t\ge1\right)\)
\(y=f\left(t\right)=t^2-t+1\)
\(minf\left(t\right)=f\left(1\right)=1\)
\(minf\left(t\right)=1\Leftrightarrow t=1\Leftrightarrow\sqrt[3]{x^2+1}=1\Leftrightarrow x=0\)
tìm x:
3^2x+1=3^x
HELP ME! Giúp mik zới ! PLEASE
TÌM x , biêt ,
a, ( 2x + 1 ) + ( 3 - x ) = 0
Help me !!!!
(2x+1)+(3-x)=0
=>2x+1=-3+x
=>2x+1-x=-3
=>x+1=-3
=>x=-3-1=-4
Vậy x=-4
a)(2x+1)+(3-x)=0
2x+1+3-x=0
x+4=0
x=-4
vậy x=-4
Help me pls:"))
Tìm đa thức B(x) thỏa mãn:A(x)=B(x).Q(x)-x+1
Biết A(x)=x^3-2x^2+x Q(x)=x-1
`@` `\text {Ans}`
`\downarrow`
Ta có:
`A(x) = B(x)* Q(x) - x + 1`
`A(x) = x^3-2x^2+x`; `Q(x) = x - 1`
`<=> B(x) * (x - 1) - x + 1 = x^3 - 2x^2 + x`
`<=> B(x) * (x - 1) = x^3 - 2x^2 + x + x - 1`
`<=> B(x) * (x - 1) = x^3 - 2x^2 + 2x - 1`
`<=> B(x) = (x^3 - 2x^2 + 2x - 1) \div (x - 1)`
`<=> B(x) = x^2 - x + 1`
Vậy, `B(x) = x^2 - x + 1.`
A(x)=B(x)*Q(x)-x+1
=>x^3-2x^2+x=B(x)(x-1)-x+1
=>B(x)*(x-1)=x^3-2x^2+x+x-1=x^3-2x^2+2x-1
=>\(B\left(x\right)=\dfrac{x^3-2x^2+2x-1}{x-1}=\dfrac{\left(x-1\right)\left(x^2+x+1\right)-2x\left(x-1\right)}{x-1}\)
=>B(x)=x^2+x+1-2x
=>B(x)=x^2-x+1
Ta có:
\(A\left(x\right)=B\left(x\right)\cdot Q\left(x\right)-x+1\)
\(\Leftrightarrow B\left(x\right)\cdot Q\left(x\right)=A\left(x\right)+x-1\)
\(\Leftrightarrow B\left(x\right)=\dfrac{A\left(x\right)+x-1}{Q\left(x\right)}\)
Mà: \(A\left(x\right)=x^3-2x^2+x\) và \(Q=x-1\) thay vào ta có:
\(\Leftrightarrow B\left(x\right)=\dfrac{x^3-2x^2+x+x-1}{x-1}\)
\(\Leftrightarrow B\left(x\right)=\dfrac{x^3-2x^2+2x-1}{x-1}\)
\(\Leftrightarrow B\left(x\right)=\dfrac{\left(x-1\right)\left(x^2-x+1\right)}{x-1}\)
\(\Leftrightarrow B\left(x\right)=x^2-x+1\)
Tìm x trong các đẳng thức:
| 2x-1 | = | 2x+3 |
| x - 1 | + 3x = 1
| 5x-3| - x = 7
Nhanh giúp với! Help me!@@ Cần gấp **
\(|5x-3|-x=7\)
\(|5x-3|=7+x\)
\(\orbr{\begin{cases}5x-3=7+x\\5x-3=-7-x\end{cases}}\)
\(\orbr{\begin{cases}5x-x=7+3\\5x+x=-7+3\end{cases}}\)
\(\orbr{\begin{cases}4x=10\\6x=-4\end{cases}}\)
\(\orbr{\begin{cases}x=2,5\\x=\frac{-2}{3}\end{cases}}\)
Vậy x = 2,5 hoặc x = -2/3
Hi Hi!
1, Tìm x ∈ Z biết
a, \(\dfrac{x-4}{15}\)=\(\dfrac{5}{3}\)
b, \(\dfrac{x}{4}\)=\(\dfrac{18}{x+1}\)
c,2x+3 ⋮ x+4
\sqrt{1} \(\dfrac{help}{me}\)
a) \(\dfrac{x-4}{15}=\dfrac{5}{3}\)
\(\Leftrightarrow x-4=15.\dfrac{5}{3}\)
\(\Leftrightarrow x-4=25\)
\(\Leftrightarrow x=29\) thỏa \(x\inℤ\)
b) \(\dfrac{x}{4}=\dfrac{18}{x+1}\left(x\ne-1\right)\)
\(\Leftrightarrow x\left(x+1\right)=18.4\)
\(\Leftrightarrow x\left(x+1\right)=72\)
vì \(72=8.9=\left(-8\right).\left(-9\right)\)
\(\Leftrightarrow x\in\left\{8;-9\right\}\left(x\inℤ\right)\)
c) \(2x+3⋮x+4\) \(\left(x\ne-4;x\inℤ\right)\)
\(\Leftrightarrow2x+3-2\left(x+4\right)⋮x+4\)
\(\Leftrightarrow2x+3-2x-8⋮x+4\)
\(\Leftrightarrow-5⋮x+4\)
\(\Leftrightarrow x+4\in\left\{-1;1;-5;5\right\}\)
\(\Leftrightarrow x\in\left\{-5;-3;-9;1\right\}\)