giải pt $\sqrt[7]{x} – \sqrt[5]{x} = \sqrt[3]{x} – \sqrt{x}$ 31/10/2021 Bởi Katherine giải pt $\sqrt[7]{x} – \sqrt[5]{x} = \sqrt[3]{x} – \sqrt{x}$
ĐKXĐ: $x≥0$ Đặt $x=y^{210}(y≥0)$ ta có PT: $\sqrt[7]{y^{210}}-\sqrt[5]{y^{210}}=\sqrt[3]{y^{210}}-\sqrt[]{y^{210}}$ $⇔y^{30}-y^{42}=y^{70}-y^{105}$ $⇔y^{105}-y^{70}-y^{42}+y^{30}=0$ $⇔y^{30}(y^{75}-y^{40}-y^{12}+1)=0$ $⇔y^{30}[y^{40}(y^{35}-1)-(y^{12}-1)]=0$ $⇔y^{30}[y^{40}(y-1)(y^{34}+y^{33}+…+1)-(y-1)(y^{11}+y^{10}+…+1)]=0$ $⇔y^{30}[(y-1)(y^{74}+y^{73}+…+y^{40})-(y-1)(y^{11}+y^{10}+…+1)]=0$ $⇔y^{30}(y-1)(y^{74}+y^{73}+…+y^{40}-y^{11}-y^{10}-…-1)=0$ Tham khảo thôi nhé!(Không khuyến khích làm theo) Bình luận
ĐKXĐ: $x≥0$
Đặt $x=y^{210}(y≥0)$ ta có PT:
$\sqrt[7]{y^{210}}-\sqrt[5]{y^{210}}=\sqrt[3]{y^{210}}-\sqrt[]{y^{210}}$
$⇔y^{30}-y^{42}=y^{70}-y^{105}$
$⇔y^{105}-y^{70}-y^{42}+y^{30}=0$
$⇔y^{30}(y^{75}-y^{40}-y^{12}+1)=0$
$⇔y^{30}[y^{40}(y^{35}-1)-(y^{12}-1)]=0$
$⇔y^{30}[y^{40}(y-1)(y^{34}+y^{33}+…+1)-(y-1)(y^{11}+y^{10}+…+1)]=0$
$⇔y^{30}[(y-1)(y^{74}+y^{73}+…+y^{40})-(y-1)(y^{11}+y^{10}+…+1)]=0$
$⇔y^{30}(y-1)(y^{74}+y^{73}+…+y^{40}-y^{11}-y^{10}-…-1)=0$
Tham khảo thôi nhé!(Không khuyến khích làm theo)