Giup minh giai bai nay voi : minh khong nhan duoc tieng viet mong moi nguoi thong cam cho minh
cau a) \(\left(x+\frac{7}{4}\right)x\frac{3}{2}=6\)
cau b) \(x:\frac{3}{5}+\frac{2}{5}=\frac{9}{5}\)
cau c) \(\frac{1}{2}:3+x=\frac{5}{3}\)
\(\frac{3}{4}x\left(\frac{8}{3}-\frac{4}{5}\right)+\frac{3}{4}-\left(\frac{4}{5}-\frac{2}{3}\right)\)
Giup minh lam bai nay voi
Mình cần giúp. Ai đó có thể giúp mình đc ko zậy
giup minh cau nay voi a. minh cam on. cho a, b la so duong
chung minh bdt \(\left(1+\frac{a}{b}\right)^5+\left(1+\frac{b}{a}\right)^{5^{ }}\ge64\)
Áp dụng BĐT AM - GM ta có :
\(\left(1+\frac{a}{b}\right)^5+\left(1+\frac{b}{a}\right)^5\ge2^5\left(\sqrt{\frac{a}{b}}\right)^5+2^5\left(\sqrt{\frac{b}{a}}\right)^5=32\left[\left(\sqrt{\frac{a}{b}}\right)^5+\left(\sqrt{\frac{b}{a}}\right)^5\right]\)
\(\ge32.2\sqrt{\left(\sqrt{\frac{a}{b}}\right)^5\left(\sqrt{\frac{b}{a}}\right)^5}=32.2=64\)(đpcm)
Dấu "=" xảy ra \(\Leftrightarrow a=b\)
a) \(\frac{\left(x+\frac{3}{4}\right)\cdot\frac{7}{2}-\frac{1}{6}}{-\left(\frac{4}{5}+\frac{1}{3}\right)\cdot\frac{1}{2}+1}=2\frac{33}{52}\)
b) \(\frac{\left(5-\frac{2}{7}\right)\cdot\frac{7}{9}:\frac{3}{5}}{\left(3x-\frac{5}{6}\right):\frac{1}{7}}=5\frac{5}{21}\)
Bài 7: Tìm x biết
CAC BAN GIUP MK NHA
AI CO CAU TRA LOI NHANH NHAT VA CHINH XAC NHAT THI MK TICK CHO
o0oNguyễno0o bn giúp mk nha bài này khó wa
a,\(x^3-\frac{4}{25}x=0\)
b,\(\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
c\(4.\left(\frac{-1}{2}\right)^3-2.\left(\frac{-1}{2}\right)^2+3.\left(\frac{-1}{2}\right)-1.\left(\frac{-1}{2}\right)^0\)
AI GIAI DUOC MINH CHO 10 TICK
a) \(x^3-\frac{4}{25}x=0\)
\(\Leftrightarrow x\left(x+\frac{2}{5}\right)\left(x-\frac{2}{5}\right)=0\)
<=> x = 0
Xét 2 trường hợp:
\(\Leftrightarrow x+\frac{2}{5}=0\)
\(x=0-\frac{2}{5}\)
\(x=-\frac{2}{5}\)
\(\Leftrightarrow x-\frac{2}{5}=0\)
\(x=0+\frac{2}{5}\)
\(x=\frac{2}{5}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm\frac{2}{5}\end{cases}}\)
b) \(\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{4}{3}\)
\(=\frac{13}{40}:\frac{4}{3}\)
\(=\frac{39}{120}=\frac{13}{40}\)
c) \(4\left(\frac{-1}{2}\right)^3-2\left(\frac{-1}{2}\right)^2+3\left(\frac{-1}{2}\right)-1\left(\frac{-1}{2}\right)^0\)
\(=4\left(\frac{-1}{2}\right)^3-2\left(\frac{-1}{2}\right)^3+3\left(\frac{-1}{2}\right)-1.1\)
\(=-\frac{1}{2}-\frac{1}{2}-\frac{3}{2}-1.1\)
\(=-\frac{5}{2}-1\)
\(=-\frac{7}{2}\)
Giup minh giai cau nay voi: ( Chi nhung ai gioi Toan lop 7 tro len moi giai duoc)
Cau 1: Cho \(\frac{1}{c}\) = \(\frac{1}{2}\) (\(\frac{1}{a}+\frac{1}{b}\) ) ( voi a, b, c khac 0 ; b khac c )
C/m rang \(\frac{a}{b}=\frac{a-c}{c-b}\)
Cau 2: Tim cac gia tri cua x,y thoa man :
\(|2x-27|^{2011}+\left(3y+10\right)^{2012}=0\)
\(\frac{1}{c}=\frac{1}{2}.\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}\Rightarrow c=\frac{2ab}{a+b}\)
\(\frac{a-c}{c-b}=\frac{a-\frac{2ab}{a+b}}{\frac{2ab}{a+b}-b}=\frac{\frac{a^2+ab-2ab}{a+b}}{\frac{2ab-ab-b^2}{a+b}}=\frac{a^2+ab-2ab}{2ab-ab-b^2}=\frac{a.\left(a-b\right)}{b.\left(a-b\right)}=\frac{a}{b}\)(ĐPCM)
\(\left|2x-27\right|^{2017}+\left(3y+10\right)^{2012}\Rightarrow\hept{\begin{cases}2x-27=0\\3y+10=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{27}{2}\\y=-\frac{10}{3}\end{cases}}\)(làm tắt nha, có gì bn thêm vào)
câu 2 : | 2x - 27 |\(^{2011}\)+ ( 3y + 10 ) \(^{2012}\)=0
=> \(\left|2x-27\right|^{2011}\)lớn hơn hoặc = 0 (1)
=> \(\left(3y+10\right)^{2012}\)>hoặc = 0(2)
mà (1) + (2) =0
nên => \(\left|2x-27\right|^{2011}=0\)và \(\left(3y+10\right)^{2012}=0\)
\(\left|2x-27\right|^{2011}=0^{2011}\) \(\left(3y+10\right)^{2012}=0^{2012}\)
\(\left|2x-27\right|=0\) 3y + 10 = 0
2x = 27 3y = -10
x = 27 : 2 y = -10 : 3
x = 13,5 y = \(\frac{-10}{3}\)
1, Ta có : \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Rightarrow\frac{1}{c}.2=\frac{1}{a}+\frac{1}{b}\)
\(\Rightarrow\frac{2}{c}=\frac{a+b}{ab}\)
\(\Rightarrow2ab=c\left(a+b\right)\)
\(\Rightarrow2ab=ac+bc\)
\(\Rightarrow2ab-ab-bc=ac+bc-ab-bc\)
\(\Rightarrow ab-bc=ac-ab\)
\(\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)
2, Ta có: \(|2x-27|^{2017}+\left(3y+10\right)^{2012}=0\)
Mà \(|2x-27|^{2017}+\left(3y+10\right)^{2012}\ge0\)
\(\Rightarrow|2x-27|^{2017}=0;\left(3y+10\right)^{2012}=0\)
\(\Rightarrow2x-27=0;3y+10=0\)
\(\Rightarrow2x=27;3y=10\)
\(\Rightarrow x=\frac{27}{2};y=\frac{10}{3}\)
a,\(\left(\frac{-1}{2}\right)^3-\left(\frac{2}{5}x+\frac{1}{3}\right)=3\)
b,\(-2\frac{1}{3}-\left(4\frac{1}{6}-\frac{4}{3}+1\frac{1}{2}\right)\)
AI GIAI DUOC MINH CHO 10 TICK.NHANH NHA MINH DANG CAN GAP
a) -1/8 -2x/5-1/3=3
-2x/5=3+1/8+1/3
-2x/5=83/24
-2x=(83×5)/24=415/24
x = (415÷-2)/24= -415/48
b) -7/3 -(25/6 -4/3+ 3/2)
= -7/3 -13/3 = -20/3
ai ma co 10 k ban chac ban phai lap chuc nink thui
\(\left[-\frac{2}{5}x^3.\left(2x-1\right)^m+\frac{2}{5}x^{m+3}\right]:\left(-\frac{2}{5}x^3\right)\)
tim x nguyen
giai ra giup minh voi
\(\left[\frac{-2}{5}x^3.\left(2x-1\right)^m+\frac{2}{5}x^{m+3}\right]:\left(\frac{-2}{5}x^3\right)\)
\(=\left[\frac{2}{5}x^3\left(2x+1\right)^m+\frac{2}{5}x^3.\left(\frac{2}{5}\right)^m\right]:\left(\frac{-2}{5}x^3\right)\)
\(=\left\{\frac{2}{5}x^3.\left[\left(2x+1\right)^m+\left(\frac{2}{5}\right)^m\right]\right\}:\left(\frac{-2}{5}x^3\right)\)
\(=\left\{\frac{2}{5}x^3.\left[2x+\frac{7}{5}\right]^m\right\}:\frac{-2}{5}x^3\)
\(=-\left(2x+\frac{7}{5}\right)^m\)
đến đây thì mình chịu
Giup minh giai bai toan nay voi
\(x\left(x+\frac{4}{y}\right)+\frac{1}{y^2}=2\)
\(x\left(2+\frac{1}{y}\right)+\frac{2}{y}=3\)
a, \(\frac{3}{5}-\frac{5}{6}\)
b, \(\left(\frac{3}{4}+\frac{-7}{2}\right).\left(\frac{2}{11}+\frac{6}{11}\right)\)
c, 25% cua A voi A=\(\frac{4}{3}.\frac{13}{9}-\frac{4}{3}.\frac{40}{9}\)
Cau 2 : Tim x bieta, x-\(\frac{3}{4}=\frac{-1}{2}\)
b, \(\frac{x+2}{3}=\frac{x-2}{2}\)
Cau 3:Lop 6A co 45 hoc sinh. So hoc sing gioi bang 20% so hoc sinh ca lop. So hoc sinh kha gap 3 lan so hoc sinh gioi. Con lai la hoc sing trung binh.a, Tinh so hoc sinh moi loai cua lop 6A
b, Tinh ti so phan tram so hoc sinh trung binh so voi so hoc sinh ca lop.