Câu 1 :
a) |x+5| + |3-x| = 8
b) 1/2 . (3/4 . x - 1/2)2020 + | 4/5 . y + 6/25| bé hơn hoặc bằng 0
c) (x-1)2 + (y+3)2 = 0
d) |x-3y|5 + |y+4| = 0
ìm x,y,z thuộc Q:
a)|x+9/2|+|y+4/3|+|z+7/2| nhỏ hơn hoặc bằng 0
b)|x+3/4|+|y-2/5|+|z+1/2| nhỏ hơn hoặc bằng 0
c) |x+19/5|+|y+1890/1975|+|z-2004|=0
d) |x+3/4|+|y-1/5|+|x+y+z|=0
a,
\(\left|x+\dfrac{9}{2}\right|\ge0\forall x\\ \left|y+\dfrac{4}{3}\right|\ge0\forall y\\ \left|z+\dfrac{7}{2}\right|\ge0\forall z\\ \Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x,y,z\)
Mà
\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\\ \Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|=0\\\left|y+\dfrac{4}{3}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\z+\dfrac{7}{2}=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-9}{2}\\y=\dfrac{-4}{3}\\z=\dfrac{-7}{2}\end{matrix}\right.\)
Vậy \(x=\dfrac{-9}{2};y=\dfrac{-4}{3};z=\dfrac{-7}{2}\)
d,
\(\left|x+\dfrac{3}{4}\right|\ge0\forall x\\ \left|y-\dfrac{1}{5}\right|\ge0\forall y\\ \left|x+y+z\right|\ge0\forall x,y,z\\ \Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x,y,z\)
Mà
\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{1}{5}=0\\x+y+z=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\x+y+z=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\\dfrac{-3}{4}+\dfrac{1}{5}+z=0\end{matrix}\right.\\\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\\dfrac{-11}{20}+z=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\z=\dfrac{11}{20}\end{matrix}\right.\)
b,
\(\left|x+\dfrac{3}{4}\right|\ge0\forall x\\ \left|y-\dfrac{2}{5}\right|\ge0\forall y\\ \left|z+\dfrac{1}{2}\right|\ge0\forall z\\ \Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\ge0\forall x,y,z\\ \)
Mà \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\le0\\ \Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{2}{5}\right|=0\\\left|z+\dfrac{1}{2}\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{2}{5}=0\\z+\dfrac{1}{2}=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{2}{5}\\z=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy ...
c,
\(\left|x+\dfrac{19}{5}\right|\ge0\forall x\\ \left|y+\dfrac{1890}{1975}\right|\ge0\forall y\\ \left|z-2004\right|\ge0\forall z\\ \Rightarrow\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|\ge0\forall x,y,z\)
Mà
\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{1890}{1975}\right|=0\\\left|z-2004\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{19}{5}=0\\y+\dfrac{1890}{1975}=0\\z-2004=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-19}{5}\\y=\dfrac{-1890}{1975}=\dfrac{-378}{395}\\z=2004\end{matrix}\right. \)
Vậy ...
Câu 1:Tính giá trị biểu thức sau:
a,(-123) + 77 + (-257) + 23 - 43
b,48 + | 48 - 147| + (-74 )
c,-2012 + ( -596) + (-201) + 496 + 301
Câu 2: Tìm x thuộc Z sao cho:
a,|-5| . | x | = | -20 |
b,| x | < -5
c,12 lớn hơn hoặc bằng | x | < 15
d,| x - 1 | + (-3) = 17
e,| x + 1 | - ( -4 ) = 5
Câu 3 : Tìm x thuộc Z sao cho
a, ( x + 1) . ( 3 - x) =0
b, ( 3x + 9 ) . ( 1 - 3x )= 0
c, | x - 5 |^5 = 32
d, | x - 7 | lớn hơn hoặc bằng 3
Câu 4 : Tìm x,y thuộc Z sao cho
a, | x + 25 | + | - y + 5 | = 0
b, | x - 1 | + | x - y + 5 | = 0
c, |x| + | y +1 | = 0
d, |x| + | y | = 1
e,( 2x - 1) . ( 4y + 2 ) = - 40
Câu 5:Tìm GTNN hoặc GTLN của biểu thức sau ( x,y thuộc Z )
a,A = | x- 3 | + 1
b, B = 3 - | x + 1 |
c, C= ( x + 1 ) ^ 2 - | 2 - y | + 11
d, D = ( x - 1 ) ^2 + | 2y + 2 | - 3
e, E = ( x+ 5 ) ^2 +(2y - 6 )^2 +1
f, F = -3 - ( 2- x )^ 2 - (3 - y )^2
Các bạn nhớ làm cả lời giải nhé !
Cảm ơn nhiều ^_^
Bạn cho nhiều thế ai làm nổi??? Từ từ từng câu 1 thôi chứ !!
Tìm x,y,z thuộc Q:
a)|x+9/2|+|y+4/3|+|z+7/2| nhỏ hơn hoặc bằng 0
b)|x+3/4|+|y-2/5|+|z+1/2| nhỏ hơn hoặc bằng 0
c) |x+19/5|+|y+1890/1975|+|z-2004|=0
d) |x+3/4|+|y-1/5|+|x+y+z|=0
Giúp mk với mn ơi
Tìm x,y,z thuộc Q:
a)|x+9/2|+|y+4/3|+|z+7/2| nhỏ hơn hoặc bằng 0
b)|x+3/4|+|y-2/5|+|z+1/2| nhỏ hơn hoặc bằng 0
c) |x+19/5|+|y+1890/1975|+|z-2004|=0
d) |x+3/4|+|y-1/5|+|x+y+z|=0
giúp mk nha mn mk đang cần gấp lắm
Hơi tắt nhá
a) Đặt \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=A\)
\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x;y;z\)
mà A\(\le0\)
\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\) phải bằng 0 đê thỏa mãn điều kiện
\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|=0\\\left|y+\dfrac{4}{3}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{9}{2}\\y=-\dfrac{4}{3}\\z=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy....
b;c)I hệt câu a nên làm tương tự nhá
d)
Hơi tắt nhá
a) Đặt \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=B\)
B=\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\x+y+z=0\end{matrix}\right.\)
Thay ra ta tính đc :\(z=-\dfrac{11}{20}\)
Vậy....
\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\)
\(\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|\ge0\\\left|y+\dfrac{4}{3}\right|\ge0\\\left|z+\dfrac{7}{2}\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\\\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\end{matrix}\right.\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|=0\Rightarrow x=-\dfrac{9}{2}\\\left|y+\dfrac{4}{3}\right|=0\Rightarrow y=-\dfrac{4}{3}\\\left|z+\dfrac{7}{2}\right|=0\Rightarrow z=-\dfrac{7}{2}\end{matrix}\right.\)
\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\le0\)
\(\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|\ge0\\\left|y-\dfrac{2}{5}\right|\ge0\\\left|z+\dfrac{1}{2}\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\ge0\\\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\le0\end{matrix}\right.\)
\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\Rightarrow x=-\dfrac{3}{4}\\\left|y-\dfrac{2}{5}\right|=0\Rightarrow y=\dfrac{2}{5}\\\left|z+\dfrac{1}{2}\right|=0\Rightarrow z=-\dfrac{1}{2}\end{matrix}\right.\)
\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\)
\(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|\ge0\\ \left|y+\dfrac{1980}{1975}\right|\ge0\\\left|z-2004\right|\ge0\end{matrix}\right.\)
\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1980}{1975}\right|+\left|z-2004\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\Rightarrow x=-\dfrac{19}{5}\\ \left|y+\dfrac{1980}{1975}\right|=0\Rightarrow y=-\dfrac{1980}{1975}\\\left|z-2004\right|=0\Rightarrow z=2004\end{matrix}\right.\)
\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)
\(\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|\ge0\\ \left|y-\dfrac{1}{5}\right|\ge0\\\left|x+y+z\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\Rightarrow x=-\dfrac{3}{4}\\\left|y-\dfrac{1}{5}\right|=0\Rightarrow y=\dfrac{1}{5}\\\left|x+y+z\right|=0\Rightarrow z+-\dfrac{11}{20}=0\Rightarrow z=\dfrac{11}{20}\end{matrix}\right.\)
Đặt \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=A\)
\(\Rightarrow A\ge0\)
Mà ĐK đề là \(A\le0\)
\(\Rightarrow A=0\)
\(\left[{}\begin{matrix}\left|x+\dfrac{3}{4}=0\right|\Rightarrow x=-\dfrac{3}{4}\\\left|y-\dfrac{2}{5}=0\right|\Rightarrow y=\dfrac{2}{5}\\\left|z+\dfrac{1}{2}=0\right|\Rightarrow z=-\dfrac{1}{2}\end{matrix}\right.\)
Các câu còn lại tương tự nhé
A giải các bất phương trình sau:
1, (x+4)/5 - x + 5 < (x+3)/3- (x-2)/2
2, (x+27)/5 - (3x-7)/4 >0
3, (7-8x)/(x^2+1) >0
4, (2x+1)/5 - (2x-2)/3 < 1
5, 1/(x+2) < 1/(x-2)
6, (x-2)/(x-5) - 3/(x-1) < 1
7, x + 6/x < 7
8, (3x-5)/x bé hơn hoặc bằng 2
9, (2x+1)/(x+1) bé hơn hoặc bằng 1
a,Cho a,b,c thỏa mãn a+b+c=0
CMR:ab+2bc+3ca bé hơn hoặc bằng 0
b, 1, CMR:(x-y)(x^4+x^3+x^2.y^2 +xy^3+y^4)=x^5-y^5
2, Cho x>y>0 và x^5+y^5=x-y
CMR: x^4+y^4<1
tìm x ,y bt (/ là giá trị tuyệt đối nhé)
a,/x-3/+/x+5/-8=0
b,/2x+1/+*2x-5/-4=0
c,/x-3/+/3x+4/+/2x-1/=8
d,/x-3y/ mũ 11 +(y+4) mũ 12=0
e,(x+y) mũ 2016 + 2017/y-1/ mũ 3 = 0
d,/x-y-5/+2015(y-3) mũ 2016=0
f,(x-1) mũ 2 + (y+3) mũ 4 = 0
g, 2(x-5) mũ 6 + 5[/2y-7/ mũ 5]=0
ch,/x=3y-1/+(3y-2) mũ 2016 =0
Nếu dc mọi người có thể chỉ rõ cho em cách giả dc ko ạ,lần sau có j em còn bt làm.Em cảm ơn ạ
Tìm x ,y ,z biet :
a, |x+3/4|+|y-1/5|+|x+y+z|=0
b, |3x-4|+|3y-5|=0
c,|x+3/4|+|y-2/5|+|z+1/2| <0
d, |x+1/5|+|3-y|=0
a) \(|x+\frac{3}{4}|+|y-\frac{1}{5}|+|x+y+z|=0\)
\(\Rightarrow|x+\frac{3}{4}|=|y-\frac{1}{5}|=|x+y+z|=0\)
\(\Rightarrow|x+\frac{3}{4}|=0\) \(\Rightarrow|y-\frac{1}{5}|=0\) \(\Rightarrow|x+y+z|=0\)
\(\Rightarrow x+\frac{3}{4}=0\) \(\Rightarrow y-\frac{1}{5}=0\) \(\Rightarrow x+y+z=0\)
\(x=\frac{-3}{4}\) \(y=\frac{1}{5}\) thay x=-3/4; y=1/5 vào biểu thức trên
ta có \(\frac{-3}{4}+\frac{1}{5}+z=0\)
\(z=0-\frac{-3}{4}-\frac{1}{5}\)
VẬY X=-3/4; Y=1/5; Z=11/20
B) \(|3x-4|+\left|3y-5\right|=0\)
\(\Rightarrow\left|3x-4\right|=\left|3y-5\right|=0\)
\(\Rightarrow\left|3x-4\right|=0\) \(\Rightarrow\left|3y-5\right|=0\)
\(3x-4=0\) \(3y-5=0\)
\(3x=4\) \(3y=5\)
\(x=\frac{4}{3}\) \(y=\frac{5}{3}\)
VẬY X= 4/3; Y=5/3
C) \(\left|x+\frac{3}{4}\right|+\left|y-\frac{2}{5}\right|+\left|z+\frac{1}{2}\right|< 0\)
ĐỂ \(\left|x+\frac{3}{4}\right|+\left|y-\frac{2}{5}\right|+\left|z+\frac{1}{2}\right|< 0\)
\(\Rightarrow\left|x+\frac{3}{4}\right|;\left|y-\frac{2}{5}\right|;\left|z+\frac{1}{2}\right|< 0\)
MÀ GIÁ TRỊ TUYỆT ĐỐI LUÔN MANG SỐ NGUYÊN DƯƠNG
\(\Rightarrow x;y;z\in\varnothing\)
d) \(\left|x+\frac{1}{5}\right|+\left|3-y\right|=0\)
\(\Rightarrow\left|x+\frac{1}{5}\right|=\left|3-y\right|=0\)
\(\Rightarrow\left|x+\frac{1}{5}\right|=0\) \(\Rightarrow\left|3-y\right|=0\)
\(x+\frac{1}{5}=0\) \(3-y=0\)
\(x=\frac{-1}{5}\) \(y=3\)
VẬY X= -1/5; Y=3
CHÚC BN HỌC TỐT!!!!!!!
Ta có :
\(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x+\frac{3}{4}=0\\y-\frac{1}{5}=0\\x+y+z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-3}{4}\\y=\frac{1}{5}\\z=0-\frac{-3}{4}-\frac{1}{5}\end{cases}}\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x=\frac{-3}{4}\\y=\frac{1}{5}\\z=\frac{11}{20}\end{cases}}\)
Vậy \(x=\frac{-3}{4};y=\frac{1}{5};z=\frac{11}{20}\)
\(b)\) Ta có :
\(\left|3x-4\right|+\left|3y-5\right|=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x-4=0\\3y-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=4\\3y=5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\y=\frac{5}{3}\end{cases}}}\)
Vậy \(x=\frac{4}{3}\) và \(y=\frac{5}{3}\)
B1: tìm các số nguyên x,y biết
a) 6/2x+1=2/7
b)24/7x+3=-4/25
c)4/x-6==y/25=-12/18
d)-1/5 nhỏ hơn hoặc bằng x/8 lớn hơn hoặc bằng 1/4
e)x+46/20=x/5/2
f)y/5/y=86/y (x/5/2;y/5/y là hỗn số)
B2: rút gọn thành phân số tối giản
a) 5^3.90.4^3/25^2.3^2.2^13
b)18.27+18.(-23)/34.4-4.52
c)15^2.16^4-15^3.16^3
d)2.3+4.6+14.21/3.5+6.10+21.35
B5: với những giá trị nguyên nào của x thì phân số sau là phân sô tối giản:
a)x-8/2x-17
b)x-4/x+1
c)10/x+7
d*)x-1/x^2
a 6/2x+1=2/7
6/2 x+1= 2/7-1
6/2 x=-5/7
x = -5/21