Microsoft Math Solver
Selesaikan
Berlatih
Unduh
Solve
Practice
Topik
Pra-Aljabar
Mean
Mode
Faktor Persekutuan Terbesar
Kelipatan Persekutuan Terkecil
Urutan Operasi
Pecahan
Pecahan Campuran
Faktorisasi Prima
Eksponen
Akar
Aljabar
Gabungkan Istilah-Istilah Serupa
Penyelesaian Satu Variabel
Faktor
Ekspansi
Menyelesaikan Pecahan
Persamaan Linear
Persamaan Kuadrat
Ketidaksetaraan
Sistem Persamaan
Matriks
Trigonometri
Menyederhanakan
Menyelesaikan
Grafik
Menyelesaikan Persamaan
Kalkulus
Turunan
Integral
Limit
Kalkulator Aljabar
Kalkulator Trigonometri
Kalkulator Kalkulus
Kalkulator Matriks
Unduh
Topik
Pra-Aljabar
Mean
Mode
Faktor Persekutuan Terbesar
Kelipatan Persekutuan Terkecil
Urutan Operasi
Pecahan
Pecahan Campuran
Faktorisasi Prima
Eksponen
Akar
Aljabar
Gabungkan Istilah-Istilah Serupa
Penyelesaian Satu Variabel
Faktor
Ekspansi
Menyelesaikan Pecahan
Persamaan Linear
Persamaan Kuadrat
Ketidaksetaraan
Sistem Persamaan
Matriks
Trigonometri
Menyederhanakan
Menyelesaikan
Grafik
Menyelesaikan Persamaan
Kalkulus
Turunan
Integral
Limit
Kalkulator Aljabar
Kalkulator Trigonometri
Kalkulator Kalkulus
Kalkulator Matriks
Selesaikan
Aljabar
trigonometri
statistik
Kalkulus
Matriks
variabel
Daftar
Evaluasi
\sqrt{\sqrt{x^{2}-5}+5}
x
2
−
5
+
5
Diferensial w.r.t. x
\frac{x}{2\sqrt{\left(\sqrt{x^{2}-5}+5\right)\left(x^{2}-5\right)}}
2
(
x
2
−
5
+
5
)
(
x
2
−
5
)
x
Grafik
Kuis
Algebra
\sqrt { \sqrt { ( \sqrt { x ^ { 2 } - 5 } ) ^ { 2 } } + 5 }
(
x
2
−
5
)
2
+
5
Soal yang Mirip dari Pencarian Web
Least value of L for \sqrt{x^2+ax}-\sqrt{x^2+bx}<L for all x>0
Least value of
L
for
x
2
+
a
x
−
x
2
+
b
x
<
L
for all
x
>
0
https://math.stackexchange.com/questions/2945698/least-value-of-l-for-sqrtx2ax-sqrtx2bxl-for-all-x0
We have \sqrt{x^2+ax}-\sqrt{x^2+bx}=\frac{(x^2+ax)-(x^2+bx)}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}=\frac{a-b}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}} for all x>0 . Because \sqrt{1+\frac{a}{x}}>1 ...
We have
x
2
+
a
x
−
x
2
+
b
x
=
x
2
+
a
x
+
x
2
+
b
x
(
x
2
+
a
x
)
−
(
x
2
+
b
x
)
=
1
+
x
a
+
1
+
x
b
a
−
b
for all
x
>
0
. Because
1
+
x
a
>
1
...
How do you solve \displaystyle\sqrt{{{3}{x}^{{{2}}}-{12}{x}+{9}}}+{3}={3}{x} ?
How do you solve
3
x
2
−
1
2
x
+
9
+
3
=
3
x
?
https://socratic.org/questions/how-do-you-solve-sqrt-3x-2-12x-9-3-3x
x = 1 only Explanation: rewrite as \displaystyle\sqrt{{{3}{x}^{{2}}-{12}{x}+{9}}} = 3x - 3 square both sides \displaystyle{3}{x}^{{2}} - 12x + 9 = \displaystyle{9}{x}^{{2}} - 18x + ...
x = 1 only Explanation: rewrite as
3
x
2
−
1
2
x
+
9
= 3x - 3 square both sides
3
x
2
- 12x + 9 =
9
x
2
- 18x + ...
Maximize the value of \sqrt{x-x^2}+\sqrt{cx-x^2} without using calculus
Maximize the value of
x
−
x
2
+
c
x
−
x
2
without using calculus
https://math.stackexchange.com/questions/2954864/maximize-the-value-of-sqrtx-x2-sqrtcx-x2-without-using-calculus
Using the Cauchy-Schwartz inequality \left(\sqrt{x-x^2}+\sqrt{c x-x^2}\right)^2\le \left(x+1-x\right)\left(x+c-x\right) = c then \sqrt{x-x^2}+\sqrt{c x-x^2}\le\sqrt{c}
Using the Cauchy-Schwartz inequality
(
x
−
x
2
+
c
x
−
x
2
)
2
≤
(
x
+
1
−
x
)
(
x
+
c
−
x
)
=
c
then
x
−
x
2
+
c
x
−
x
2
≤
c
How do you find the domain of \displaystyle{h}{\left({x}\right)}=\sqrt{{{4}-{x}}}+\sqrt{{{x}^{{2}}-{1}}} ?
How do you find the domain of
h
(
x
)
=
4
−
x
+
x
2
−
1
?
https://socratic.org/questions/how-do-you-find-the-domain-of-h-x-sqrt-4-x-sqrt-x-2-1
Domain of \displaystyle{h}{\left({x}\right)} \displaystyle={\left(-\infty,-{1}\right]}\cup{\left[{1},{4}\right]} Explanation: Domain is all possible allowable \displaystyle{x} value ...
Domain of
h
(
x
)
=
(
−
∞
,
−
1
]
∪
[
1
,
4
]
Explanation: Domain is all possible allowable
x
value ...
Factoring the expression (\sqrt{x^2} -a)^2 + M = 0
Factoring the expression
(
x
2
−
a
)
2
+
M
=
0
https://math.stackexchange.com/q/1542043
Taking \sqrt{-M} is problematic. You need to establish that M is >= 0 first. From the comments you seem to think that since M represents "any number of terms" you can skip this step... this is ...
Taking
−
M
is problematic. You need to establish that
M
is >= 0 first. From the comments you seem to think that since
M
represents "any number of terms" you can skip this step... this is ...
Why does this outcome change depending on infinity
Why does this outcome change depending on infinity
https://math.stackexchange.com/questions/659725/why-does-this-outcome-change-depending-on-infinity
No, you're missing an absolute value sign on x in the denominator. Put differently, note that for large values of x, \sqrt{x^2+2x} \approx |x| + \text{sgn}(x) and \sqrt{x^2-2x} \approx |x| - \text{sgn}(x). ...
No, you're missing an absolute value sign on
x
in the denominator. Put differently, note that for large values of
x
,
x
2
+
2
x
≈
∣
x
∣
+
sgn
(
x
)
and
x
2
−
2
x
≈
∣
x
∣
−
sgn
(
x
)
.
...
Lebih banyak Item
Bagikan
Salin
Disalin ke clipboard
Contoh
Persamaan kuadrat
{ x } ^ { 2 } - 4 x - 5 = 0
x
2
−
4
x
−
5
=
0
Trigonometri
4 \sin \theta \cos \theta = 2 \sin \theta
4
sin
θ
cos
θ
=
2
sin
θ
Persamaan linear
y = 3x + 4
y
=
3
x
+
4
Aritmetika
699 * 533
6
9
9
∗
5
3
3
Matriks
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
[
2
5
3
4
]
[
2
−
1
0
1
3
5
]
Persamaan simultan
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
{
8
x
+
2
y
=
4
6
7
x
+
3
y
=
4
7
Diferensial
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
d
x
d
(
x
−
5
)
(
3
x
2
−
2
)
Integral
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
∫
0
1
x
e
−
x
2
d
x
Limit
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}
x
→
−
3
lim
x
2
+
2
x
−
3
x
2
−
9
Kembali ke atas