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Giải:
Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=k\Rightarrow\left\{\begin{matrix}a=2k\\b=3k\\c=4k\end{matrix}\right.\)
Ta có: \(a^2+3b^2-2c^2=\left(-16\right)\)
\(\Rightarrow4k^2+27k^2-32k^2=-16\)
\(\Rightarrow\left(-1\right)k^2=-16\)
\(\Rightarrow k^2=16\)
\(\Rightarrow k=\pm4\)
+) \(k=4\Rightarrow a=8;b=12;c=16\)
+) \(k=-4\Rightarrow a=-8;b=-12;c=-16\)
Vậy bộ số \(\left(x;y;z\right)\) là \(\left(8;12;16\right);\left(-8;-12;-16\right)\)
Bạn có làm trong này rồi nhé Câu hỏi của Phạm Vũ Ngọc Duy
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\\ \Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\\ =\dfrac{a^2+3b^2-2c^2}{4+27-32}=-\dfrac{16}{-1}=16\\ \Rightarrow a=\pm8;b=\pm12;c=\pm16\)
Giải:
Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\Rightarrow\left\{{}\begin{matrix}a=2k\\b=3k\\c=4k\end{matrix}\right.\)
Ta có: \(a^2+3b^2-2c^2=-16\)
\(\Rightarrow4k^2+27k^2-32k^2=-16\)
\(\Rightarrow-k^2=-16\)
\(\Rightarrow k^2=16\)
\(\Rightarrow k=\pm4\)
+) \(k=4\Rightarrow a=8,b=12,c=16\)
+) \(k=-4\Rightarrow a=-8;b=-12;c=-16\)
Vậy bộ số \(\left(a;b;c\right)\) là \(\left(8;12;16\right);\left(-8;-12;-16\right)\)
\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a^2}{4}=\dfrac{3b^2}{27}=\dfrac{2c^2}{32}=\dfrac{a^2+3b^2-2c^2}{4+27-32}=\dfrac{-16}{-1}=16\)
\(\Rightarrow a^2=64,b^2=144,c^2=256\) hay:
\(\left(a;b;c\right)=\left(8;12;16\right)=\left(-8;-12;-16\right)\)
ĐS: \(\left(a;b;c\right)=\left(8;12;16\right)=\left(-8;-12;-16\right)\)
a, Vì \(\dfrac{a}{c}=\dfrac{c}{b}\Rightarrow ab=c^2\)
Ta có :
\(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{\left(b+a\right)\left(b-a\right)}{a^2+ab}=\dfrac{\left(b+a\right)\left(b-a\right)}{a\left(a+b\right)}=\dfrac{b-a}{a}\)
Vậy \(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b-a}{a}\)
a)đặt \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=k\(\Rightarrow\)a=bk, c=dk
\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (1)
\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (2)
từ (1),(2)\(\Rightarrow\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)
b)ta có:
\(\dfrac{ab}{cd}=\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}\)
câu c bn tự giải nhé dễ mak ahihihi
chúc bn hc tốt
a) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có:
\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (1)
\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (2)
Từ (1) và (2) suy ra \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)
b) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=q\Rightarrow\left\{{}\begin{matrix}a=bq\\c=dq\end{matrix}\right.\)
Ta có:
\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bq+b}{dq+d}\right)^2=\left[\dfrac{b\left(q+1\right)}{d\left(q+1\right)}\right]^2=\dfrac{b}{d}\) (1)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bq\right)^2+b^2}{\left(dq\right)^2+d^2}=\dfrac{b^2.q^2+b^2}{d^2.q^2+d^2}=\dfrac{b^2\left(q^2+1\right)}{d^2\left(q^2+1\right)}=\dfrac{b}{d}\) (2)
Từ (1) và (2) suy ra \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\) => \(\dfrac{a}{c}=\dfrac{b}{d}\)
áp dụng tính chất dãy tỉ số = nhau ta có
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a+3b}{2c+3d}=\dfrac{2a-3b}{2c-3d}\)
= \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\) (đpcm)
a) Ta có:
+) a/2=b/3
=>a=2b/3
+) b/5=c/4
=>c=4b/5
Lại có:
a-b+c=49
=> 2b/3 -b + 4b/5 =49
=> 7b/15==49
=> b= 105
Khi đó:
+) a=2b/3=2.105/3=70
+)c=4b/5=4.105/5=84
Vậy a=70; b=105; c=84...
chúc bạn học tốt
Từ giả thiết \(\Rightarrow a=4a';b=4b';c=4c'\)
Nên \(\dfrac{a+b+c}{a'+b'+c'}=\dfrac{4\left(a'+b'+c'\right)}{a'+b'+c'}=4\)
\(\dfrac{a-3b+2c}{a'-3b'+2c'}=\dfrac{4\left(a'-3b'+2c'\right)}{a'-3b'+2c'}=4\)
@Phạm Ngân Hà mk ko biet cach nay co dung ko ban xem giup mk nhe :v
\(\dfrac{b}{b'}=\dfrac{3b}{3b'};\dfrac{c}{c'}=\dfrac{2c}{2c'}\)
de bai: \(\dfrac{a}{a'}=\dfrac{b}{b'}=\dfrac{c}{c'}\Leftrightarrow\dfrac{a}{a'}=\dfrac{3b}{3b'}=\dfrac{2c}{2c'}=\dfrac{a-3b+2c}{a'-3b'+2c'}=4\)(TCDTSBN)
Xét \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\)
\(\Rightarrow a=2k;b=3k;c=4k\) (1)
Thay (1) vào \(a^2+3b^2-2c^2\)
\(\Rightarrow\)\(a^2+3b^2-2c^2\)\(=\left(2k\right)^2+3\left(3k\right)^2-2\left(4k\right)^2\)
\(=4k^2+27k^2-32k^2=-k^2=-64\)
\(\Rightarrow k^2=64\Rightarrow\left[{}\begin{matrix}k=-8\\k=8\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=\pm16\\b=\pm24\\c=\pm32\end{matrix}\right.\)
Đặt a/2=b/3=c/4=k
=>a=2k; b=3k; c=4k
Ta có: \(a^2+3b^2-2c^2=-16\)
\(\Leftrightarrow4k^2+27k^2-32k^2=-16\)
\(\Leftrightarrow k^2=16\)
Trường hợp 1: k=4
=>a=8; b=12; c=16
Trường hợp 2: k=-4
=>a=-8; b=-12; c=-16
cảm ơn