5. 判断下列计算是否正确。正确的在(
(1) $\frac{6}{7} × ( \frac{7}{6} + 2.8 ) = 1 + 2.8 = 3.8$ (
改正:
(2) $42 - 35 × \frac{4}{7} = 7 × \frac{4}{7} = 4$ ()
改正:
×
)里画“√”,错误的画“×”,并改正。(1) $\frac{6}{7} × ( \frac{7}{6} + 2.8 ) = 1 + 2.8 = 3.8$ (
×
)改正:
(2) $42 - 35 × \frac{4}{7} = 7 × \frac{4}{7} = 4$ ()
改正:
答案
5. (1) × 改正:$\frac{6}{7}×(\frac{7}{6}+2.8)=\frac{6}{7}×\frac{7}{6}+\frac{6}{7}×2.8=1+2.4=3.4$
(2) × 改正:$42-35×\frac{4}{7}=42-20=22$
(2) × 改正:$42-35×\frac{4}{7}=42-20=22$
6. 花江峡谷大桥位于贵州黔西南州与安顺市交界处,巍然屹立于被誉为“地球裂缝”的花江大峡谷之上,大桥全长 2890 m,是目前世界山区峡谷第一大跨度钢桁梁悬索桥。花江峡谷大桥桥面距水面的垂直高度为 625 m,主跨径比垂直高度的$\frac{11}{5}$还要多 45 m。花江峡谷大桥的主跨径为多少米?
答案
$625×\frac{11}{5}+45=1420$(m)
7. 观察下面的算式并找出规律。
$1×\frac{1}{2}=1-\frac{1}{2}$ $\frac{1}{2}×\frac{1}{3}=\frac{1}{2}-\frac{1}{3}$ $\frac{1}{3}×\frac{1}{4}=\frac{1}{3}-\frac{1}{4}$ ……
(1)$\frac{1}{11}-\frac{1}{12}=(\quad)×(\quad)$
(2)利用发现的规律,计算下面各题。
$1×\frac{1}{2}+\frac{1}{2}×\frac{1}{3}+\frac{1}{3}×\frac{1}{4}+\frac{1}{4}×\frac{1}{5}$

$1×\frac{1}{2}=1-\frac{1}{2}$ $\frac{1}{2}×\frac{1}{3}=\frac{1}{2}-\frac{1}{3}$ $\frac{1}{3}×\frac{1}{4}=\frac{1}{3}-\frac{1}{4}$ ……
(1)$\frac{1}{11}-\frac{1}{12}=(\quad)×(\quad)$
(2)利用发现的规律,计算下面各题。
$1×\frac{1}{2}+\frac{1}{2}×\frac{1}{3}+\frac{1}{3}×\frac{1}{4}+\frac{1}{4}×\frac{1}{5}$
答案
(1) $\frac{1}{11}$ $\frac{1}{12}$
(2) $1×\frac{1}{2}+\frac{1}{2}×\frac{1}{3}+\frac{1}{3}×\frac{1}{4}+\frac{1}{4}×\frac{1}{5}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}$
$=1-\frac{1}{5}$
$=\frac{4}{5}$
$1×\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+…+\frac{1}{2024×2025}+\frac{1}{2025×2026}$
$=1×\frac{1}{2}+\frac{1}{2}×\frac{1}{3}+\frac{1}{3}×\frac{1}{4}+…+\frac{1}{2024}×\frac{1}{2025}+\frac{1}{2025}×\frac{1}{2026}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+…+\frac{1}{2024}-\frac{1}{2025}+\frac{1}{2025}-\frac{1}{2026}$
$=1-\frac{1}{2026}$
$=\frac{2025}{2026}$
(2) $1×\frac{1}{2}+\frac{1}{2}×\frac{1}{3}+\frac{1}{3}×\frac{1}{4}+\frac{1}{4}×\frac{1}{5}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}$
$=1-\frac{1}{5}$
$=\frac{4}{5}$
$1×\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+…+\frac{1}{2024×2025}+\frac{1}{2025×2026}$
$=1×\frac{1}{2}+\frac{1}{2}×\frac{1}{3}+\frac{1}{3}×\frac{1}{4}+…+\frac{1}{2024}×\frac{1}{2025}+\frac{1}{2025}×\frac{1}{2026}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+…+\frac{1}{2024}-\frac{1}{2025}+\frac{1}{2025}-\frac{1}{2026}$
$=1-\frac{1}{2026}$
$=\frac{2025}{2026}$
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