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1) Phân số đầu nhân 2.
_ Phân số thứ 2 nhân 3, p/s thứ 3 giữ nguyên.
_ Lấy phân số đầu + p/s thứ 2 - p/s thứ 3.
_ Dựa vào dãy tỉ số bằng nhau tìm x, y, z.
2) \(x-y-z=0\Rightarrow x=y+z\)
Khi đó thay vào B được:
\(B=\left(1-\dfrac{z}{y+z}\right)\left(1-\dfrac{y+z}{y}\right)\left(1+\dfrac{y}{z}\right)\)
\(=\dfrac{y}{y+z}.\dfrac{z}{y}.\dfrac{y+z}{z}\)
\(=1\)
Vậy B = 1.
\(a,\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Rightarrow\left(x-1\right)^{x+2}\left[1-\left(x-1\right)^2\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x\left(2-x\right)=0\end{cases}}}\)
=> x=1 ; x=0 ; x=2
Vậy..
Bài 1 :
b) \(\left|x-3\right|=5\)
\(\Rightarrow\orbr{\begin{cases}x-3=-5\\x-3=5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}\)
Vậy x thuộc {-2; 8}
c) \(\left|2x+1\right|=x-8\)
\(\Rightarrow\orbr{\begin{cases}2x+1=-x+8\\2x+1=x-8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=7\\x=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=-9\end{cases}}\)
Vậy x thuộc {-9; 7/3}
Câu c) tớ không chắc, thông cảm.
=))
\(3\left(x-1\right)=2\left(y-2\right)\Rightarrow\frac{x-1}{2}=\frac{y-2}{3}\)(1)
\(4\left(y-2\right)=3\left(z-3\right)\Rightarrow\frac{y-2}{3}=\frac{z-3}{4}\)(2)
Từ (1) và (2) suy ra \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
\(\Leftrightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-x+3}{4+9-4}=\frac{45}{9}=5\)
\(\Rightarrow\hept{\begin{cases}x=\left(5.4+2\right):2=11\\y=\left(5.9+6\right):3=17\\z=\left(4.5+3\right)=23\end{cases}}\)
a, \(\left|3x-4\right|+\left|3y+5\right|=0\)
Ta có :
\(\left|3x-4\right|\ge0\forall x;\left|3y+5\right|\ge0\forall x\\ \)
\(\Rightarrow\left|3x-4\right|+\left|3y+5\right|\ge0\forall x\\ \Rightarrow\left\{{}\begin{matrix}3x-4=0\\3y+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=4\\3y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=-\dfrac{5}{3}\end{matrix}\right.\\ Vậy.........\)
b, \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\)
Ta có :
\(\left|x+\dfrac{19}{5}\right|\ge0\forall x;\left|y+\dfrac{1890}{1975}\right|\ge0\forall y;\left|z-2004\right|\ge0\forall z \)
\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{19}{5}=0\\y+\dfrac{1890}{1975}=0\\z-2004=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{19}{5}\\y=-\dfrac{1890}{1975}\\z=2004\end{matrix}\right.\\ Vậy............\)
c, \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\)
Ta có : \(\left|x+\dfrac{9}{2}\right|\ge0\forall x;\left|y+\dfrac{4}{3}\right|\ge0\forall y;\left|z+\dfrac{7}{2}\right|\ge0\forall z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\z+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{9}{2}\\y=-\dfrac{4}{3}\\z=-\dfrac{7}{2}\end{matrix}\right.\\ Vậy............\)
d, \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)
Ta có :
\(\left|x+\dfrac{3}{4}\right|\ge0\forall x;\left|y-\dfrac{1}{5}\right|\ge0\forall y;\left|x+y+z\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{1}{5}=0\\x+y+z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\z=0-\dfrac{1}{5}+\dfrac{3}{4}=\dfrac{11}{20}\end{matrix}\right.\\ Vậy.......\)
e, Câu cuối bn làm tương tự như câu a, b, c nhé!
Bài 2:
\(\left\{{}\begin{matrix}\left(2x-\dfrac{1}{2}\right)^2\ge0\\\left(y+\dfrac{1}{2}\right)^2\ge0\\\left(z-\dfrac{1}{3}\right)^2\ge0\end{matrix}\right.\Rightarrow\left(2x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2+\left(z-\dfrac{1}{3}\right)^2\ge0\)Mà \(\left(2x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2+\left(z-\dfrac{1}{3}\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(2x-\dfrac{1}{2}\right)^2=0\\\left(y+\dfrac{1}{2}\right)^2=0\\\left(z-\dfrac{1}{3}\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=\dfrac{-1}{2}\\z=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{4},y=\dfrac{-1}{2},z=\dfrac{1}{3}\)
1)
a) \(2x+\dfrac{5}{2}=\dfrac{7}{2}\)
\(\Leftrightarrow2x=\dfrac{7}{2}-\dfrac{5}{2}\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
b) \(\left|5-\dfrac{1}{2}x\right|=\left|-\dfrac{1}{5}\right|\)
\(\Leftrightarrow\left|5-\dfrac{1}{2}x\right|=\dfrac{1}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}5-\dfrac{1}{2}x=\dfrac{1}{5}\\5-\dfrac{1}{2}x=-\dfrac{1}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{48}{5}\\x=\dfrac{52}{5}\end{matrix}\right.\)
Vậy \(x_1=\dfrac{48}{5};x_2=\dfrac{52}{5}\)
Bài 1
Hình thì bạn tự vẽ nha (chú ý cứ vẽ △ABC vuông cân )
Ta có AB=0,5 BC →AB =\(\dfrac{1}{2}\) BC hay 2AB = BC
+, 2AB=BC →2(AB2 )=BC2 (1)
Xét △ABC có góc A vuông (gt)
➝AB2 + AC2 =BC2 (Định lí Pi-ta-go) (2)
Từ (1) và (2) → AB2 + AC2 =2(AB2 )
→ AC2 = 2(AB2 ) - AB2 →AC2 = AB2 hay AB = AC Xét △ABC có AB = AC (cmt) và góc A vuông →△ABC là tam giác vuông cân → góc B = góc C = 45oMình viết nhầm bạn sửa 2(AB2) thành (2AB)2 nhé ...!!!!