e^(3x)=8则x加5分之3x等于7分之6于

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相关电视台3x^2+10x-8=0求x等于多少
由3x²+10x-8=0得﹙x+4﹚﹙3x-2﹚= 0∴x=-4或x=2/3
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求根公式啊。。。。x=-4或2/3
(3x+2)(x-4)=0 x=-2/3或x=4 ∵a=3,b=-10,c=-8 b -4ac=﹙﹣10﹚ ﹣4×3×﹙﹣8﹚=196﹥0 ∴X1=﹙10+√196﹚/﹙2
扫描下载二维码如图,将边长为4的正方形置于平面直角坐标系第一象限,使AB边落在x轴正半轴上,且A点的坐标是(1,0).(1)直线$y=\frac{4}{3}x-\frac{8}{3}$经过点C,且与x轴交于点E,求四边形AECD的面积;(2)若直线l经过点E,且将正方形ABCD分成面积相等的两部分,求直线l的解析式;(3)若直线l1经过点F($-\frac{3}{2},0$)且与直线y=3x平行.将(2)中直线l沿着y轴向上平移1个单位,交x轴于点M,交直线l1于点N,求△NMF的面积.
(1)先求出E点的坐标,根据梯形的面积公式即可求出四边形AECD的面积;(2)根据已知求出直线1上点G的坐标,设直线l的解析式是y=kx+b,把E、G的坐标代入即可求出解析式;(3)根据直线l1经过点F($-\frac{3}{2},0$)且与直线y=3x平行,知k=3,把F的坐标代入即可求出b的值即可得出直线11,同理求出解析式y=2x-3,进一步求出M、N的坐标,利用三角形的面积公式即可求出△MNF的面积.
(1)$y=\frac{4}{3}x-\frac{8}{3}$,当y=0时,x=2,∴E(2,0),由已知可得:AD=AB=BC=DC=4,AB∥DC,∴四边形AECD是梯形,∴四边形AECD的面积S=$\frac{1}{2}$×(2-1+4)×4=10,答:四边形AECD的面积是10.(2)在DC上取一点G,使CG=AE=1,则St梯形AEGD=S梯形EBCG,∴G点的坐标为(4,4),设直线l的解析式是y=kx+b,代入得:$\left\{\begin{array}{l}{4=4k+b}\\{0=2k+b}\end{array}\right.$,解得:$\left\{\begin{array}{l}{k=2}\\{b=-4}\end{array}\right.$,即:y=2x-4,答:直线l的解析式是y=2x-4.(3)∵直线l1经过点F($-\frac{3}{2},0$)且与直线y=3x平行,设直线11的解析式是y1=kx+b,则:k=3,代入得:0=3×(-$\frac{3}{2}$)+b,解得:b=$\frac{9}{2}$,∴y1=3x+$\frac{9}{2}$已知将(2)中直线l沿着y轴向上平移1个单位,则所得的直线的解析式是y=2x-4+1,即:y=2x-3,当y=0时,x=$\frac{3}{2}$,∴M($\frac{3}{2}$,0),解方程组$\left\{\begin{array}{l}{y=3x+\frac{9}{2}}\\{y=2x-3}\end{array}\right.$得:$\left\{\begin{array}{l}{x=-\frac{15}{2}}\\{y=-18}\end{array}\right.$,即:N(-$\frac{15}{2}$,-18),S△NMF=$\frac{1}{2}$×[$\frac{3}{2}$-(-$\frac{3}{2}$)]×|-18|=27.答:△NMF的面积是27.

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