làm giúp mìn bài 4 và 5 với

Bài 4: Ta có: 1) $\frac{2a+3c}{2b+3d}=\frac{2a+3\times \frac{b}{a}\times a}{2b+3\times \frac{b}{a}\times b}=\frac{2+\frac{3b}{a}}{2+\frac{3b}{a}}=1$ $\frac{2a-3c}{2b-3d}=\frac{2a-3\times \frac{b}{a}\times a}{2b-3\times \frac{b}{a}\times b}=\frac{2-\frac{3b}{a}}{2-\frac{3b}{a}}=1$ Vậy $\frac{2a+3c}{2b+3d}=\frac{2a-3c}{2b-3d}$ 2) $\frac{2a+5b}{3a-4b}=\frac{2a+5\times \frac{a}{c}\times c}{3a-4\times \frac{a}{c}\times c}=\frac{2+5\times \frac{a}{c}}{3-4\times \frac{a}{c}}$ $\frac{2c+5d}{3c-4d}=\frac{2c+5\times \frac{c}{a}\times a}{3c-4\times \frac{c}{a}\times a}=\frac{2+5\times \frac{c}{a}}{3-4\times \frac{c}{a}}$ Vậy $\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}$ 3) $\frac{3a+5b}{6a-7b}=\frac{3a+5\times \frac{a}{c}\times c}{6a-7\times \frac{a}{c}\times c}=\frac{3+5\times \frac{a}{c}}{6-7\times \frac{a}{c}}$ $\frac{3c+5d}{6c-7d}=\frac{3c+5\times \frac{c}{a}\times a}{6c-7\times \frac{c}{a}\times a}=\frac{3+5\times \frac{c}{a}}{6-7\times \frac{c}{a}}$ Vậy $\frac{3a+5b}{6a-7b}=\frac{3c+5d}{6c-7d}$ 4) $\frac{11a+3b}{11c+3d}=\frac{11a+3\times \frac{a}{c}\times c}{11c+3\times \frac{c}{a}\times a}=\frac{11+3\times \frac{a}{c}}{11+3\times \frac{c}{a}}$ $\frac{3a-11b}{3c-11d}=\frac{3a-11\times \frac{a}{c}\times c}{3c-11\times \frac{c}{a}\times a}=\frac{3-11\times \frac{a}{c}}{3-11\times \frac{c}{a}}$ Vậy $\frac{11a+3b}{11c+3d}=\frac{3a-11b}{3c-11d}$ 5) $\frac{7a-4b}{3a+5b}=\frac{7a-4\times \frac{a}{c}\times c}{3a+5\times \frac{a}{c}\times c}=\frac{7-4\times \frac{a}{c}}{3+5\times \frac{a}{c}}$ $\frac{7c-4d}{3c+5d}=\frac{7c-4\times \frac{c}{a}\times a}{3c+5\times \frac{c}{a}\times a}=\frac{7-4\times \frac{c}{a}}{3+5\times \frac{c}{a}}$ Vậy $\frac{7a-4b}{3a+5b}=\frac{7c-4d}{3c+5d}$ 6) $\frac{3a+5b}{2a-7b}=\frac{3a+5\times \frac{a}{c}\times c}{2a-7\times \frac{a}{c}\times c}=\frac{3+5\times \frac{a}{c}}{2-7\times \frac{a}{c}}$ $\frac{3c+5d}{2c-7d}=\frac{3c+5\times \frac{c}{a}\times a}{2c-7\times \frac{c}{a}\times a}=\frac{3+5\times \frac{c}{a}}{2-7\times \frac{c}{a}}$ Vậy $\frac{3a+5b}{2a-7b}=\frac{3c+5d}{2c-7d}$ Bài 5: 1) Ta có $\frac ab=\frac cd$. Do đó $\frac{a-c}c=\frac{a}{c}-\frac cc=\frac{a}{c}-1=\frac{d}{b}-1=\frac{d-b}{b}=\frac{b-d}d$ 2) Ta có $\frac ab=\frac cd$. Do đó $\frac{a-b}a=\frac aa-\frac ba=1-\frac ba=1-\frac dc=\frac{c-d}c$ 3) Ta có $\frac ab=\frac cd$. Do đó $\frac a{a+b}=\frac{\frac ab}{\frac ab+1}=\frac{\frac cd}{\frac cd+1}=\frac c{c+d}$ 4) Ta có $\frac ab=\frac cd$. Do đó $\frac{a+c}c=\frac ac+1=\frac db+1=\frac{b+d}d$ 5) Ta có $\frac ab=\frac cd$. Do đó $\frac{a+b}b=\frac ab+1=\frac cd+1=\frac{c+d}d$