如果\|begin(array)(cc)a&b c&dend(array)|=ad-bc,那么当a=\5/3,b=3,c=\3/4,d=2\1/4时,\|begin(array)(cc)a&b c&dend(array)|= \3/2. 相关知识点: 试题来源: 解析 原式利用题中的新定义计算即可. 解:∵\|begin(array)(cc)a&b c&dend(array)|=ad-bc,a=\5/3,b=3...
【解析】∵ Veft|\begin{array}{cc}a&b\\_c&d\end{array}\right.\rig $$ = a - b + c - d $$ ∴$$\left\{ \begin{matrix} x y - 3 x ^ { 2 } - 2 x y - x ^ { 2 } \end{matrix} \right.$$ $$ - 2 x ^ { 2 } - 3 - 5 + x y \\ = ( x y - 3 x ^ ...
\end{array} \right. 1. 2. 3. 4. 5. y = { a 12345 b + x 78 l 910 y = \left\{ \begin{array}{ll} a & 12345\\ b+x & 78\\ l &910 \end{array} \right.y=⎩⎨⎧ab+xl1234578910 c:center y = \left\{ \begin{array}{cc} a & 12345\\ b+x & 78\\...
r:right y=\left\{ \begin{array}{rr} a&12345\\ b+x&78\\ l&910 \end{array} \right. 1. 2. 3. 4. 5. l:left y=\left\{ \begin{array}{ll} a&12345\\ b+x&78\\ l&910 \end{array} \right. 1. 2. 3. 4. 5. c:center y=\left\{ \begin{array}{cc} a&12345\\ b+x...
5.现规定一种新的运算$\left|\begin{array}{cc}a&b\\ c&d\end{array}\right|$=ad-bc.那么$\left|\begin{array}{cc}2&3\\ 2-x&4\end{array}\right|$=9时.x=$\frac{7}{3}$.
【题目】我们定义Neft|Nbegin{array}{cc}a&bll c&d\end{array}right.rig=ad-bc,例如eft|begin{array}{
规定四个数a、b、c、d的一种运算法则如下:\left|\begin{array}{cc}a&b\\ c&d\end{array}\right.\right|=ad-cb,计算\left|\begin{array}{cc}\frac{1}{2}&2\\ \frac{1}{3}&\frac{4}{3}\end{array}\right.\right|= 0. 相关知识点: 有理数 有理数的运算 有理数的应用 新定义问题 ...
2.对于任意实数a.b.c.d.我们规定$\left|\begin{array}{cc}a&b\\ c&d\end{array}\right|$=ad-bc.若-8<$\left|\begin{array}{cc}x-1&x+1\\ x&x+5\end{array}\right|$<4.求整数x的值.
我按照在线教程进行操作,效果很好。当我将相同的代码应用于我自己的端点时,我得到了这个异常:java.lang.IllegalStateException: Expected BEGIN_ARRAY but was BEGIN_OBJECT at line 1 column 2 path $我不知道如何解决这个问题。 界面: public interface MyApiService {...
19.对于任意实数a.b.c.d.我们将式子$\left|\begin{array}{cc}a&b\\ c&d\end{array}\right|$称为二阶行列式.并且规定$\left|\begin{array}{cc}a&b\\ c&d\end{array}\right|=ad-bc$.(1)计算$\left|\begin{array}{cc}2×{10}^{7}&3×{10}^{6}\\ 4×{10}^{6}&7×{10}^{5}\end{