96 - 3x = 3. 32
Tìm x:
a) (3x-7)^5=32
b) (4x-1)^3=27.125
c) 2^x+2^2^x+1=96
d)3^4x +4=81 ^x+3
Giair giùm nha
a) (3x-7)5=32
=> (3x-7)5=25
=> 3x-7=2
=> 3x=2+7=9
=>x=9:3=3
b) (4x-1)3=27.125
=> (4x-1)3=33.53
=> (4x-1)3=(3.5)3
=> (4x-1)3=153
=> 4x-1=15
(Các bước còn lại tương tự câu a)
a)(3x-7)^5=32
(3x-7)^5=2^5
3x-7=2
3x=2+7
3x=9
x=9:3
x=3
2x+22x+1=96
2x+22x.21=96
2x+22x.2=96
2x+22x=96:2
2x+22x=48
96*2-96*9/15+16*32/3-16+24*35=?
tim x
2 mũ x+2 mũ x+1 =96
(3x+2)mũ 3=2.32
5 mũ x+1 -5x =500
1 x 32+7x8+4x
=200x
Tính
a,\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
b,\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
c,\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)
d,\(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}\)
\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}-2\sqrt{2}\right)^2}\)
\(=3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)
\(=3-2\sqrt{2}+3+2\sqrt{2}=6\)
\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\sqrt{\left(5-2\sqrt{6}\right)^2}+\sqrt{\left(5+2\sqrt{6}\right)^2}\)
\(=5-2\sqrt{6}+5+2\sqrt{6}=10\)
\(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}+\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)
\(=2\sqrt{2}-\sqrt{5}+3\sqrt{5}+2\sqrt{2}=2\sqrt{5}+4\sqrt{2}\)
a: \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
\(=3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)
b: \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(=3-2\sqrt{2}+3+2\sqrt{2}\)
=6
c: Ta có: \(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)
\(=5-2\sqrt{6}+5+2\sqrt{6}\)
=10
d: Ta có: \(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{13-4\sqrt{10}}+\sqrt{53+4\sqrt{90}}\)
\(=2\sqrt{2}-\sqrt{5}+3\sqrt{5}+2\sqrt{2}\)
\(=2\sqrt{5}+4\sqrt{2}\)
Tính
a) 22+36= 96−32= 62−30=
89–47= 44+44= 45−5=
b) 32+3−2= 56−20−4= 23+14−15=
a) 22+36=58 96−32=64 62−30=32
89−47=42 44+44=88 45−5=40
tính nhanh
96 * 37 - 3 * 26 * 32 - 96
mình sẽ tích cho ai có câu trả lời đúng nhanh và đầy đủ
96 * 37 - 3 * 26 * 32 - 96
= 96 * 37 - 96 * 26 - 96 * 1
= 96 * ( 37 - 26 - 1 - 1 )
= 96 * 10
= 960
=
96 x 37 - 3 x 26 x 32 - 96
= 3552- 3 x 26 x 32 - 96
= 3552 - 78 x 32 -96
= 3552 - 2496 -96
= 1056 - 96
= 960
ủng hộ nhé
ai k mình mình k lại
96 x 37 - 3 x 26 x 32 - 96
= 96 x 37 - 96 x 26 - 96 x 1
=96 x (27-26- 1)
=96 x 0
=0
96 : 24 = ?
90 : 18 = ?
84 : 14 = ?
96 : 32 = ?
96:24=4
90:18=5
84:14=7
96:32=3
Tính
a, \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
b,\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
c, \(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)
a, \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
= \(\sqrt{3^2-2.3.\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{6^2-2.6.\sqrt{6}+\left(\sqrt{6}\right)^2}\)
= \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(6-\sqrt{6}\right)^2}\)
= \(\left|3-\sqrt{6}\right|+\left|6-\sqrt{6}\right|\)
= \(3-\sqrt{6}+6-\sqrt{6}\)
= \(9-2\sqrt{6}\)
b. Đặt B = \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
Nhận xét : B > 0 , bình phương hai vế ta được :
\(B^2=\left(\sqrt{17-3\sqrt{32}}\right)^2+\left(\sqrt{17+3\sqrt{32}}\right)^2\)
\(B^2=17-3\sqrt{32}+17+3\sqrt{32}+2\sqrt{\left(17-3\sqrt{32}\right)\left(17+3\sqrt{32}\right)}\)
\(B^2=34+2\sqrt{17^2-\left(3\sqrt{32}\right)^2}\)
\(B^2=34+2\sqrt{289-288}\)
\(B^2=34+2=36\)
=> \(B=\pm\sqrt{36}\) mà B > 0 nên \(B=\sqrt{36}=6\)
c, Đặt C = \(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)
Nhận xét : C > 0 , bình phương hai vế ta đươc :
\(C^2=\left(\sqrt{49-5\sqrt{96}}\right)^2+\left(\sqrt{49+5\sqrt{96}}\right)^2\)
\(C^2=49-5\sqrt{96}+49+5\sqrt{96}+2\sqrt{\left(49-5\sqrt{96}\right)\left(49+5\sqrt{96}\right)}\)
\(C^2=98+2\sqrt{49^2-\left(5\sqrt{96}\right)^2}\)
\(C^2=98+2\sqrt{2401-2400}\)
\(C^2=98+2=100\)
=> \(C=\pm\sqrt{100}\) mà C > 0 nên \(C=\sqrt{100}=10\)
a) Ta có: \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
\(=\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{27-2\cdot3\sqrt{3}\cdot2\sqrt{2}+8}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}-2\sqrt{2}\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|3\sqrt{3}-2\sqrt{2}\right|\)
\(=3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)(Vì \(\left\{{}\begin{matrix}3>\sqrt{6}\\3\sqrt{3}>2\sqrt{2}\end{matrix}\right.\))
b) Ta có: \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(=\frac{\sqrt{34-6\sqrt{32}}+\sqrt{34+6\sqrt{32}}}{\sqrt{2}}\)
\(=\frac{\sqrt{18-2\cdot3\sqrt{2}\cdot4+16}+\sqrt{18+2\cdot3\sqrt{2}\cdot4+16}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(3\sqrt{2}-4\right)^2}+\sqrt{\left(3\sqrt{2}+4\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|3\sqrt{2}-4\right|+\left|3\sqrt{2}+4\right|}{\sqrt{2}}\)
\(=\frac{3\sqrt{2}-4+3\sqrt{2}+4}{\sqrt{2}}\)(Vì \(3\sqrt{2}>4>0\))
\(=\frac{6\sqrt{2}}{\sqrt{2}}=6\)
tính nhanh
96 * 37 - 3 * 26 * 32 - 96
minh se tích cho ai có câu trả lời nhanh và đúng
Tính theo mẫu :
32 dam x 3 = 96dam
25 m x 2 =
15 km x 4 =
34 cm x 6 =
96 cam : 3 = 32 cm.
36 hm : 3 =
70 km : 7 =
55 dm : 5 =
25 m x 2 = 50 m
15 km x 4 = 60 km
34 cm x 6 = 204 cm
36 hm : 3 = 12 hm
70 km : 7 = 10 km
55 dm : 5 = 11dm
25m x2 =50 m
15km x4 =60 km
34 cm x6= 204cm
96cm :3 =32 cm
36hm :3 =12 hm
70km :7 =10 km
55dm :5 =11 dm